Chemical equation and limiting reagent Questions in English

(B) Step $1$: Calculate moles of $H^+$ and $OH^-$.
$n_{H^+} = 2 \times \text{Molarity} \times \text{Volume (L)} = 2 \times 0.2 \times 0.4 = 0.16 \ mol$.
$n_{OH^-} = 1 \times \text{Molarity} \times \text{Volume (L)} = 0.1 \times 0.6 = 0.06 \ mol$.
Step $2$: Identify the limiting reagent. Since $n_{OH^-} < n_{H^+}$,$OH^-$ is the limiting reagent.
Step $3$: Calculate heat released $(q)$.
$q = n_{OH^-} \times |\Delta_{r}H| = 0.06 \ mol \times 57.1 \times 10^3 \ J \ mol^{-1} = 3426 \ J$.
Step $4$: Calculate temperature change $(\Delta T)$.
Total mass of solution = $(400 + 600) \ mL \times 1.0 \ g \ mL^{-1} = 1000 \ g$.
$q = m \times c \times \Delta T \Rightarrow 3426 = 1000 \times 4.18 \times \Delta T$.
$\Delta T = \frac{3426}{4180} \approx 0.8196 \ K$.
Expressing in $10^{-2} \ K$: $0.8196 \times 100 \times 10^{-2} \ K = 81.96 \times 10^{-2} \ K$.
Rounding to the nearest integer,we get $82 \times 10^{-2} \ K$.

(A) The combustion reaction is: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Moles of $CH_4 = \frac{100 \, g}{16 \, g/mol} = 6.25 \, mol$
Moles of $O_2 = \frac{208 \, g}{32 \, g/mol} = 6.5 \, mol$
To find the limiting reagent,calculate the ratio of moles to stoichiometric coefficient:
For $CH_4: \frac{6.25}{1} = 6.25$
For $O_2: \frac{6.5}{2} = 3.25$
Since the ratio for $O_2$ is smaller,$O_2$ is the limiting reagent.
From the stoichiometry,$2 \, mol$ of $O_2$ produces $1 \, mol$ of $CO_2$.
Therefore,$6.5 \, mol$ of $O_2$ will produce $\frac{6.5}{2} = 3.25 \, mol$ of $CO_2$.
Mass of $CO_2 = 3.25 \, mol \times 44 \, g/mol = 143 \, g$.

(C) $1$. Calculate the molar mass of $Fe_{3}O_{4} = (3 \times 56) + (4 \times 16) = 168 + 64 = 232 \ g \ mol^{-1}$.
$2$. Calculate the moles of $Fe_{3}O_{4} = \frac{4.640 \times 10^{3} \ g}{232 \ g \ mol^{-1}} = 20 \ mol$.
$3$. Calculate the molar mass of $CO = 12 + 16 = 28 \ g \ mol^{-1}$.
$4$. Calculate the moles of $CO = \frac{2.520 \times 10^{3} \ g}{28 \ g \ mol^{-1}} = 90 \ mol$.
$5$. According to the stoichiometry,$1 \ mol$ of $Fe_{3}O_{4}$ requires $4 \ mol$ of $CO$. Thus,$20 \ mol$ of $Fe_{3}O_{4}$ requires $20 \times 4 = 80 \ mol$ of $CO$.
$6$. Since we have $90 \ mol$ of $CO$ (which is more than $80 \ mol$),$Fe_{3}O_{4}$ is the limiting reagent.
$7$. From the equation,$1 \ mol$ of $Fe_{3}O_{4}$ produces $3 \ mol$ of $Fe$. Therefore,$20 \ mol$ of $Fe_{3}O_{4}$ produces $20 \times 3 = 60 \ mol$ of $Fe$.
$8$. Mass of $Fe$ produced $= 60 \ mol \times 56 \ g \ mol^{-1} = 3360 \ g$.

(D) The balanced chemical equation is: $X + Y + 3 Z \rightarrow XYZ_3$.
Initial moles: $X = 1 \ mol$,$Y = 1 \ mol$,$Z = 0.05 \ mol$.
To find the Limiting Reagent ($L$.$R$.),divide the moles by their stoichiometric coefficients:
$X: 1/1 = 1$
$Y: 1/1 = 1$
$Z: 0.05/3 \approx 0.0167$
Since $Z$ has the smallest ratio,$Z$ is the Limiting Reagent.
Moles of $XYZ_3$ formed = $\frac{1}{3} \times \text{moles of } Z = \frac{0.05}{3} \ mol$.
Molar mass of $XYZ_3 = 10 + 20 + (3 \times 30) = 120 \ g/mol$.
Yield of $XYZ_3 = \frac{0.05}{3} \times 120 = 0.05 \times 40 = 2 \ g$.

(D) The balanced chemical equation is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
Initial moles of $H_2SO_4 = 2 \ L \times 0.2 \ M = 0.4 \ mol$.
Initial moles of $NaOH = 2 \ L \times 0.1 \ M = 0.2 \ mol$.
Since $NaOH$ is the limiting reagent,$0.2 \ mol$ of $NaOH$ will react with $0.1 \ mol$ of $H_2SO_4$ to produce $0.1 \ mol$ of $Na_2SO_4$.
Total volume of the solution = $2 \ L + 2 \ L = 4 \ L$.
Molarity of $Na_2SO_4 = \frac{0.1 \ mol}{4 \ L} = 0.025 \ M$.
$0.025 \ M = 25 \ mM$.

(C) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Calculate the number of moles of reactants:
$n(N_{2}) = \frac{20 \ g}{28 \ g/mol} \approx 0.714 \ mol$
$n(H_{2}) = \frac{5 \ g}{2 \ g/mol} = 2.5 \ mol$
Determine the limiting reagent by dividing moles by the stoichiometric coefficient:
For $N_{2}: \frac{0.714}{1} = 0.714$
For $H_{2}: \frac{2.5}{3} \approx 0.833$
Since $0.714 < 0.833$,$N_{2}$ is the limiting reagent.
The amount of $NH_{3}$ produced depends on the limiting reagent:
$n(NH_{3}) = 2 \times n(N_{2}) = 2 \times 0.714 = 1.428 \ mol \approx 1.42 \ mol$.

(B) At $STP$,$1 \, L$ of water gas contains $1:1$ ratio of $CO$ and $H_2$ gases.
Thus,$V_{CO} = 0.5 \, L$ and $V_{H_2} = 0.5 \, L$.
Volume of $O_2$ in $9 \, L$ of air $= 9 \times 0.20 = 1.8 \, L$.
Upon ignition,$CO$ reacts with $O_2$ according to the equation: $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)}$.
Since $CO$ is the limiting reagent ($0.5 \, L$ of $CO$ requires $0.25 \, L$ of $O_2$),the volume of $CO_2$ produced is equal to the volume of $CO$ consumed,which is $0.5 \, L$.
Number of moles of $CO_2 = \frac{\text{Volume at } STP}{22.4 \, L/mol} = \frac{0.5}{22.4} \approx 0.022 \, mol$.

(B) The combustion reaction is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
Volume of $O_2$ present in $750 \, L$ of air $= \frac{20}{100} \times 750 \, L = 150 \, L$.
According to the stoichiometry,$1 \, L$ of $CH_4$ requires $2 \, L$ of $O_2$.
Therefore,$50 \, L$ of $CH_4$ requires $50 \times 2 = 100 \, L$ of $O_2$.
Remaining volume of $O_2 = 150 \, L - 100 \, L = 50 \, L$.
Since $22.4 \, L$ of any gas at $STP$ corresponds to $1 \, mole$,the number of moles of $O_2$ remaining is $\frac{50}{22.4} \approx 2.23 \, moles$.
The value closest to this is $2.2 \, moles$.

(B) The correct option is $(b)$.
When baking soda $(NaHCO_3)$ is mixed with vinegar (which contains acetic acid,$CH_3COOH$),a chemical reaction occurs that produces sodium acetate,water,and carbon dioxide gas.
The balanced chemical equation is:
$CH_3COOH_{(aq)} + NaHCO_{3(s)} \longrightarrow CH_3COONa_{(aq)} + H_2O_{(l)} + CO_{2(g)}$
Thus,the gas released is carbon dioxide $(CO_2)$.

(B) The balanced chemical equation for the reaction is: $N_2 + 3H_2 \rightleftharpoons 2NH_3$.
Given: $1$ mole of $N_2$ and $3.01 \times 10^{23}$ molecules of $H_2$.
Since $1$ mole of $H_2$ contains $6.022 \times 10^{23}$ molecules,the number of moles of $H_2$ is $\frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.5$ mole.
According to the stoichiometry,$3$ moles of $H_2$ produce $2$ moles of $NH_3$.
Therefore,$0.5$ mole of $H_2$ will produce $\frac{2}{3} \times 0.5 = \frac{1}{3}$ mole of $NH_3$.
Number of molecules of $NH_3 = \frac{1}{3} \times 6.022 \times 10^{23} \approx 2.0 \times 10^{23}$ molecules.
Thus,the correct option is $B$.

(C) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$
At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Moles of $H_{2(g)} = \frac{22.4 \ L}{22.4 \ L/mol} = 1 \ mol$.
Moles of $Cl_{2(g)} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \ mol$.
Since $Cl_{2(g)}$ is the limiting reagent,it will be completely consumed.
According to the stoichiometry,$1 \ mol$ of $Cl_{2(g)}$ produces $2 \ mol$ of $HCl_{(g)}$.
Therefore,$0.25 \ mol$ of $Cl_{2(g)}$ will produce $0.25 \times 2 = 0.5 \ mol$ of $HCl_{(g)}$.

(C) The balanced chemical equation is: $2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$
Moles of $C = \frac{12 \ g}{12 \ g \ mol^{-1}} = 1 \ mol$.
Moles of $O_2 = \frac{48 \ g}{32 \ g \ mol^{-1}} = 1.5 \ mol$.
According to the stoichiometry,$2 \ mol$ of $C$ requires $1 \ mol$ of $O_2$. Therefore,$1 \ mol$ of $C$ requires $0.5 \ mol$ of $O_2$.
Since we have $1.5 \ mol$ of $O_2$,$C$ is the limiting reagent.
From the equation,$2 \ mol$ of $C$ produces $2 \ mol$ of $CO$,so $1 \ mol$ of $C$ produces $1 \ mol$ of $CO$.
Volume of $1 \ mol$ of $CO$ at $STP$ = $22.7 \ L = 227 \times 10^{-1} \ L$.

(D) The balanced chemical equation for the reaction is:
$3 BaCl_2 + 2 Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6 NaCl$
From the stoichiometry of the reaction:
$3$ moles of $BaCl_2$ react with $2$ moles of $Na_3PO_4$ to produce $1$ mole of $Ba_3(PO_4)_2$.
Given:
$5$ moles of $BaCl_2$ and $2$ moles of $Na_3PO_4$.
To find the limiting reagent:
For $2$ moles of $Na_3PO_4$,we require $(3/2) \times 2 = 3$ moles of $BaCl_2$.
Since we have $5$ moles of $BaCl_2$ (which is more than $3$ moles),$BaCl_2$ is in excess and $Na_3PO_4$ is the limiting reagent.
Calculation of product:
$2$ moles of $Na_3PO_4$ produce $1$ mole of $Ba_3(PO_4)_2$.
Therefore,the maximum number of moles of $Ba_3(PO_4)_2$ formed is $1$.

(A) The balanced chemical equation for the reaction is:
$P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 2S_2Cl_2 + 4SO_2$
Comparing this with the given reaction:
$P_4 + 8SOCl_2 \rightarrow 4A + xSO_2 + 2B$
We can identify:
$A = PCl_3$
$B = S_2Cl_2$
$x = 4$
Therefore,the correct values are $PCl_3$,$S_2Cl_2$ and $4$.

(A) The balanced chemical equation is: $3 PbCl_2 + 2 (NH_4)_3 PO_4 \rightarrow Pb_3(PO_4)_2 + 6 NH_4 Cl$
Calculate the moles required for each reactant:
For $PbCl_2$: $72 \ mmol / 3 = 24 \ mmol$
For $(NH_4)_3 PO_4$: $50 \ mmol / 2 = 25 \ mmol$
Since $24 < 25$,$PbCl_2$ is the limiting reagent.
The amount of $Pb_3(PO_4)_2$ formed is determined by the limiting reagent:
$mmol \text{ of } Pb_3(PO_4)_2 = \frac{1}{3} \times mmol \text{ of } PbCl_2 = \frac{72}{3} = 24 \ mmol$.

(A) The chemical reaction is: $NaOH + HCl \longrightarrow NaCl + H_2O$
First,calculate the mass of $HCl$ in $25 \ mL$ of $0.75 \ M$ solution:
$Moles \ of \ HCl = Molarity \times Volume(L) = 0.75 \times 0.025 = 0.01875 \ mol$
Mass of $HCl = Moles \times Molar \ Mass = 0.01875 \times 36.5 \approx 0.684375 \ g$
According to the stoichiometry,$1 \ mol$ of $HCl$ reacts with $1 \ mol$ of $NaOH$ $(40 \ g)$:
$0.01875 \ mol \ HCl \ reacts \ with \ 0.01875 \ mol \ NaOH$
Mass of $NaOH$ reacted $= 0.01875 \times 40 = 0.75 \ g$
Mass of $NaOH$ left unreacted $= 1 \ g - 0.75 \ g = 0.25 \ g$
Since $1 \ g = 1000 \ mg$,the mass left is $0.25 \times 1000 = 250 \ mg$.

(B) The balanced chemical equation is: $2 \ Al(s) + 3 \ H_2SO_4(aq) \longrightarrow Al_2(SO_4)_3(aq) + 3 \ H_2(g)$
Moles of $Al = \frac{5.4 \ g}{27.0 \ g \ mol^{-1}} = 0.2 \ mol$
Moles of $H_2SO_4 = M \times V(L) = 5.0 \ mol \ L^{-1} \times 0.050 \ L = 0.25 \ mol$
According to the stoichiometry,$2 \ mol$ of $Al$ requires $3 \ mol$ of $H_2SO_4$.
For $0.2 \ mol$ of $Al$,required $H_2SO_4 = \frac{3}{2} \times 0.2 = 0.3 \ mol$.
Since we have only $0.25 \ mol$ of $H_2SO_4$,$H_2SO_4$ is the limiting reagent.
Moles of $H_2$ produced $= \frac{3}{3} \times 0.25 \ mol = 0.25 \ mol$.
Using the ideal gas law $PV = nRT$:
$V = \frac{nRT}{P} = \frac{0.25 \ mol \times 0.082 \ atm \ L \ mol^{-1} \ K^{-1} \times 300 \ K}{1.0 \ atm} = 6.15 \ L$.

(C) The balanced chemical equation is: $4Al + 3O_2 \longrightarrow 2Al_2O_3$.
Moles of $Al = \frac{81.0 \ g}{27.0 \ g \ mol^{-1}} = 3.0 \ mol$.
Moles of $O_2 = \frac{128.0 \ g}{32.0 \ g \ mol^{-1}} = 4.0 \ mol$.
According to the stoichiometry,$4 \ mol$ of $Al$ requires $3 \ mol$ of $O_2$.
For $3.0 \ mol$ of $Al$,required $O_2 = \frac{3}{4} \times 3.0 = 2.25 \ mol$.
Since $4.0 \ mol$ of $O_2$ is available,$Al$ is the limiting reagent.
Moles of $Al_2O_3$ produced $= \frac{2}{4} \times 3.0 = 1.5 \ mol$.
Molar mass of $Al_2O_3 = (2 \times 27.0) + (3 \times 16.0) = 54.0 + 48.0 = 102.0 \ g \ mol^{-1}$.
Mass of $Al_2O_3 = 1.5 \ mol \times 102.0 \ g \ mol^{-1} = 153.0 \ g$.

(A) Calculate the moles of reactants:
$Moles \ of \ Fe_3O_4 = \frac{2.32 \times 10^3 \times 10^3 \ g}{232 \ g \ mol^{-1}} = 10^4 \ mol$.
$Moles \ of \ CO = \frac{2.8 \times 10^2 \times 10^3 \ g}{28 \ g \ mol^{-1}} = 10^4 \ mol$.
From the balanced equation $Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$,the stoichiometric ratio is $1:4$.
Since we have $10^4 \ mol$ of $Fe_3O_4$ and $10^4 \ mol$ of $CO$,$CO$ is the limiting reagent because it requires $4 \times 10^4 \ mol$ of $CO$ to react completely with $10^4 \ mol$ of $Fe_3O_4$.
$Moles \ of \ Fe \ produced = \frac{3}{4} \times (moles \ of \ CO) = \frac{3}{4} \times 10^4 = 7500 \ mol$.
$Mass \ of \ Fe = 7500 \ mol \times 56 \ g \ mol^{-1} = 420000 \ g = 420 \ kg$.
Therefore,$x = 420$.

(C) The balanced chemical equation is: $CaCO_{3(s)} + 2 HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g \ mol^{-1}$.
Moles of $CaCO_3 = \frac{1000 \ g}{100 \ g \ mol^{-1}} = 10 \ mol$.
Moles of $HCl = Molarity \times Volume (L) = 0.76 \ mol \ L^{-1} \times 0.250 \ L = 0.19 \ mol$.
According to the stoichiometry,$2 \ mol$ of $HCl$ reacts with $1 \ mol$ of $CaCO_3$.
Since $0.19 \ mol$ of $HCl$ is present,it will react with $0.095 \ mol$ of $CaCO_3$.
Thus,$HCl$ is the Limiting Reagent $(L.R.)$.
Moles of $CaCl_2$ formed $= \frac{1}{2} \times \text{moles of } HCl = \frac{0.19}{2} = 0.095 \ mol$.
Molar mass of $CaCl_2 = 40 + (2 \times 35.5) = 111 \ g \ mol^{-1}$.
Mass of $CaCl_2 = 0.095 \ mol \times 111 \ g \ mol^{-1} = 10.545 \ g$.

(C) The balanced chemical equation is: $C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4 CO_2 + 5 H_2O$.
Molar mass of $C_4H_{10} = 4 \times 12 + 10 \times 1 = 58 \ g \ mol^{-1}$.
Molar mass of $O_2 = 2 \times 16 = 32 \ g \ mol^{-1}$.
Moles of $C_4H_{10} = \frac{174 \times 10^3 \ g}{58 \ g \ mol^{-1}} = 3000 \ mol$.
Moles of $O_2 = \frac{320 \times 10^3 \ g}{32 \ g \ mol^{-1}} = 10000 \ mol$.
According to the stoichiometry,$1 \ mol$ of $C_4H_{10}$ requires $6.5 \ mol$ of $O_2$.
For $3000 \ mol$ of $C_4H_{10}$,$O_2$ required $= 3000 \times 6.5 = 19500 \ mol$.
Since only $10000 \ mol$ of $O_2$ is available,$O_2$ is the limiting reagent.
Moles of $H_2O$ formed $= 5 \times (\frac{10000}{6.5}) = 7692.3 \ mol$.
Mass of $H_2O = 7692.3 \ mol \times 18 \ g \ mol^{-1} = 138461.5 \ g$.
Since density $= 1 \ g \ mL^{-1}$,volume $= 138461.5 \ mL = 138.46 \ L$.
The nearest integer is $138$.

(D) The balanced chemical equation is $A + 2B \rightarrow 3C$.
According to the stoichiometry,$1$ mole of $A$ reacts with $2$ moles of $B$ to produce $3$ moles of $C$.
Given moles of $A = 20$ and moles of $B = 14$.
To find the limiting reagent,calculate the ratio of moles to stoichiometric coefficient:
For $A$: $20 / 1 = 20$.
For $B$: $14 / 2 = 7$.
Since the ratio for $B$ is smaller,$B$ is the limiting reagent.
The amount of $C$ produced depends on the limiting reagent $B$.
From the equation,$2$ moles of $B$ produce $3$ moles of $C$.
Therefore,$14$ moles of $B$ will produce $(3 / 2) \times 14 = 21$ moles of $C$.

(C) Let the weight of both $X$ and $Y$ be $w \ g$.
Number of moles of $X = \frac{w}{36}$.
Number of moles of $Y = \frac{w}{24}$.
The balanced chemical equation is $2X + 3Y \rightarrow X_2Y_3$.
According to the stoichiometry,$2 \text{ moles of } X$ react with $3 \text{ moles of } Y$.
For $1 \text{ mole of } X$,we need $\frac{3}{2} = 1.5 \text{ moles of } Y$.
For $\frac{w}{36} \text{ moles of } X$,we need $\frac{w}{36} \times 1.5 = \frac{w}{24} \text{ moles of } Y$.
Since the available moles of $Y$ are exactly equal to the required moles of $Y$,neither reactant is in excess.
Therefore,no reactant is left over.

(A) $1$. Identify the limiting reagent: According to the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,$1 \ mol$ of $N_2$ requires $3 \ mol$ of $H_2$.
$2$. For $0.5 \ mol$ of $N_2$,the required $H_2$ is $0.5 \times 3 = 1.5 \ mol$.
$3$. Since only $0.5 \ mol$ of $H_2$ is available,$H_2$ is the limiting reagent.
$4$. Calculate theoretical yield: $3 \ mol$ of $H_2$ produces $2 \ mol$ of $NH_3$. Therefore,$0.5 \ mol$ of $H_2$ produces $(2/3) \times 0.5 = 0.333 \ mol$ of $NH_3$.
$5$. Calculate percentage yield: $\text{Percentage yield} = (\text{Actual yield} / \text{Theoretical yield}) \times 100 = (0.25 / 0.333) \times 100 \approx 75 \%$.

(A) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
Given,heat of combustion of $CH_4 = -100 \ kJ/mol$.
Molar mass of $CH_4 = 16 \ g/mol$ and $O_2 = 32 \ g/mol$.
Number of moles of $CH_4 = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$.
Number of moles of $O_2 = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$.
According to the stoichiometry,$1 \ mol$ of $CH_4$ requires $2 \ mol$ of $O_2$. Since we have only $1 \ mol$ of $O_2$,$O_2$ is the limiting reagent.
The heat evolved depends on the limiting reagent $(O_2)$.
For $2 \ mol$ of $O_2$,heat evolved is $100 \ kJ$.
For $1 \ mol$ of $O_2$,heat evolved = $\frac{100 \ kJ}{2} = 50 \ kJ$.

(D) $n_{H_2} = \frac{64}{2} = 32 \ mol$
$n_{O_2} = \frac{64}{32} = 2 \ mol$
According to the reaction $2H_2 + O_2 \rightarrow 2H_2O$,$1 \ mol$ of $O_2$ requires $2 \ mol$ of $H_2$.
Therefore,$2 \ mol$ of $O_2$ requires $4 \ mol$ of $H_2$.
Since $H_2$ is present in excess $(32 \ mol > 4 \ mol)$,$O_2$ is the limiting reagent.
$A$. Incorrect.
$B$. Correct,$O_2$ is the limiting reagent.
$C$. $2 \ mol$ of $O_2$ produces $4 \ mol$ of $H_2O = 4 \times 18 = 72 \ g \ H_2O$.
Moles of $H_2$ left $= 32 - 4 = 28 \ mol$.
Mass of $H_2$ left $= 28 \ mol \times 2 \ g/mol = 56 \ g$.
Thus,the mixture contains $72 \ g \ H_2O$ and $56 \ g \ H_2$. Correct.

(A) The limiting reagent is the reactant that is entirely consumed in a chemical reaction.
Since it is completely consumed,it determines the extent of the reaction and consequently limits the maximum amount of product that can be formed.
Therefore,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(C) The balanced chemical equation is: $C + O_2 \longrightarrow CO_2$
Moles of $C = \frac{4 \ g}{12 \ g/mol} = \frac{1}{3} \ mol$
Moles of $O_2 = \frac{4 \ g}{32 \ g/mol} = \frac{1}{8} \ mol$
Since $1 \ mol$ of $C$ reacts with $1 \ mol$ of $O_2$,$O_2$ is the limiting reagent because $\frac{1}{8} < \frac{1}{3}$.
The amount of $CO_2$ produced depends on the limiting reagent $(O_2)$.
From the stoichiometry,$1 \ mol$ of $O_2$ produces $1 \ mol$ of $CO_2$.
Therefore,$\frac{1}{8} \ mol$ of $O_2$ will produce $\frac{1}{8} \ mol$ of $CO_2$.

(A) The chemical reaction between methyl magnesium iodide and water is given by:
$CH_3MgI + H_2O \rightarrow CH_4 + MgI(OH)$
From the balanced chemical equation,$1$ mole of methyl magnesium iodide reacts with $1$ mole of water to produce $1$ mole of methane.
Therefore,to prepare $n$ moles of methane from $n$ moles of methyl magnesium iodide,$n$ moles of water molecules are required.

(D) The balanced chemical equation for the combustion of carbon is: $C(s) + O_2(g) \rightarrow CO_2(g)$.
Given amount of carbon = $2 \,mol$.
Given mass of dioxygen $(O_2)$ = $16 \,g$.
Molar mass of $O_2$ = $32 \,g/mol$.
Number of moles of $O_2$ = $\frac{16 \,g}{32 \,g/mol} = 0.5 \,mol$.
According to the stoichiometry,$1 \,mol$ of $C$ reacts with $1 \,mol$ of $O_2$ to produce $1 \,mol$ of $CO_2$.
Since we have $2 \,mol$ of $C$ and only $0.5 \,mol$ of $O_2$,$O_2$ is the Limiting Reagent $(LR)$.
The amount of $CO_2$ produced depends on the amount of $LR$ $(O_2)$.
Thus,$0.5 \,mol$ of $O_2$ will produce $0.5 \,mol$ of $CO_2$.
Mass of $CO_2$ produced = $0.5 \,mol \times 44 \,g/mol = 22 \,g$.

(D) The reaction is $C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g)$.
$1$ mole of $C_2H_4$ reacts with $1$ mole of $HCl$ to produce $1$ mole of $C_2H_5Cl$.
Given: $V(C_2H_4) = 200 \ mL$,$V(HCl) = 150 \ mL$.
Since $HCl$ is the limiting reagent,$150 \ mL$ of $HCl$ reacts with $150 \ mL$ of $C_2H_4$ to produce $150 \ mL$ of $C_2H_5Cl$.
Initial volume $V_1 = 150 \ mL (C_2H_4) + 150 \ mL (HCl) = 300 \ mL = 0.3 \ dm^3$.
Final volume $V_2 = 150 \ mL (C_2H_5Cl) = 0.15 \ dm^3$.
Work $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
$W = -1 \ bar \times (0.15 \ dm^3 - 0.3 \ dm^3) = -1 \ bar \times (-0.15 \ dm^3) = 0.15 \ dm^3 \ bar$.
Since $1 \ dm^3 \ bar = 100 \ J$,
$W = 0.15 \times 100 \ J = 15 \ J$.

(B) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$
Moles of $H_2 = \frac{0.4 \text{ g}}{2 \text{ g/mol}} = 0.2 \text{ mol}$
Moles of $Cl_2 = \frac{7.4 \text{ g}}{71 \text{ g/mol}} \approx 0.1042 \text{ mol}$
Since $1 \text{ mol}$ of $H_2$ reacts with $1 \text{ mol}$ of $Cl_2$, $Cl_2$ is the limiting reagent because it is present in a smaller amount.
Moles of $HCl$ formed $= 2 \times \text{moles of } Cl_2 = 2 \times 0.1042 = 0.2084 \text{ mol}$
At $273 \text{ K}$ and $1 \text{ bar}$ ($STP$ conditions), the molar volume of an ideal gas is approximately $22.7 \text{ L/mol}$.
Volume of $HCl = 0.2084 \text{ mol} \times 22.7 \text{ L/mol} \approx 4.73 \text{ L}$.
Given the options, $4.67 \text{ L}$ is the closest value, which is obtained by using the older $STP$ definition $(22.4 \text{ L/mol})$.
Thus, the correct option is $(b)$.

(A) The balanced chemical equation is:
$NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$
Molar mass of $AgNO_3 = 108 + 14 + (3 \times 16) = 170 \ g/mol$
Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$
Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$
Given amounts:
$NaCl = 11.70 \ g$
$AgNO_3 = 3.4 \ g$
From the stoichiometry,$1 \ mol$ of $AgNO_3$ reacts with $1 \ mol$ of $NaCl$.
$170 \ g$ of $AgNO_3$ requires $58.5 \ g$ of $NaCl$.
$3.4 \ g$ of $AgNO_3$ requires $\frac{58.5}{170} \times 3.4 = 1.17 \ g$ of $NaCl$.
Since we have $11.70 \ g$ of $NaCl$ (which is more than $1.17 \ g$),$AgNO_3$ is the limiting reagent.
The amount of $AgCl$ produced depends on the limiting reagent $(AgNO_3)$:
$170 \ g$ of $AgNO_3$ produces $143.5 \ g$ of $AgCl$.
$3.4 \ g$ of $AgNO_3$ produces $\frac{143.5}{170} \times 3.4 = 2.87 \ g$ of $AgCl$.
Thus,the mass of $AgCl$ precipitated is $2.87 \ g$.

(B) The balanced chemical equation is: $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$.
According to the stoichiometry,$48 \ g$ of $Mg$ reacts with $32 \ g$ of $O_2$.
For $1.0 \ g$ of $Mg$,the required amount of $O_2$ is $\frac{32 \ g \ O_2}{48 \ g \ Mg} \times 1.0 \ g \ Mg = 0.667 \ g \ O_2$.
Since only $0.28 \ g$ of $O_2$ is available,$O_2$ is the limiting reagent and $Mg$ is in excess.
Now,calculate the amount of $Mg$ consumed by $0.28 \ g$ of $O_2$: $\frac{48 \ g \ Mg}{32 \ g \ O_2} \times 0.28 \ g \ O_2 = 0.42 \ g \ Mg$.
The amount of $Mg$ left in excess is $1.0 \ g - 0.42 \ g = 0.58 \ g$.

(A) The neutralization reaction is: $HCl + NaOH \rightarrow NaCl + H_{2}O$
Initial moles of $HCl = \frac{500 \times 0.1}{1000} = 0.05 \text{ mol}$
Initial moles of $NaOH = \frac{200 \times 0.2}{1000} = 0.04 \text{ mol}$
Since $NaOH$ is the limiting reagent,$0.04 \text{ mol}$ of $HCl$ will react with $0.04 \text{ mol}$ of $NaOH$.
The heat of neutralization for $1 \text{ mole}$ of a strong acid and a strong base is $57.3 \text{ kJ/mol}$.
Therefore,the heat evolved for $0.04 \text{ mol}$ is $57.3 \times 0.04 \text{ kJ} = 2.292 \text{ kJ}$.

(D) The reaction for the formation of baking soda $(NaHCO_3)$ from sodium carbonate $(Na_2CO_3)$ is given by the equation:
$Na_2CO_3 + H_2O + CO_2 \rightarrow 2NaHCO_3$
From the balanced chemical equation,it is clear that $1 \text{ mole}$ of $H_2O$ reacts with $1 \text{ mole}$ of $CO_2$.
Therefore,the molar ratio of $H_2O : CO_2$ is $1:1$.

(D) The given chemical reaction is the thermal decomposition of lead$(II, IV)$ oxide $(Pb_3O_4)$.
To balance the equation $x Pb_3O_4 \rightarrow y PbO + O_2$:
$1$. Balance the lead $(Pb)$ atoms: On the left,there are $3x$ atoms of $Pb$. On the right,there are $y$ atoms of $Pb$. So,$3x = y$.
$2$. Balance the oxygen $(O)$ atoms: On the left,there are $4x$ atoms of $O$. On the right,there are $y + 2$ atoms of $O$.
$3$. Substitute $y = 3x$ into the oxygen balance equation: $4x = 3x + 2$.
$4$. Solving for $x$: $4x - 3x = 2$,which gives $x = 2$.
$5$. Now find $y$: $y = 3x = 3(2) = 6$.
Thus,the balanced equation is $2 Pb_3O_4 \rightarrow 6 PbO + O_2$.

(A) The balanced chemical equation is: $AgNO_{3(aq)} + NaCl_{(aq)} \rightarrow AgCl_{(s)} + NaNO_{3(aq)}$
Step $1$: Calculate the mass of reactants.
Mass of $AgNO_3 = 25 \ mL \times 0.35 = 8.75 \ g$.
Moles of $AgNO_3 = \frac{8.75 \ g}{169.87 \ g/mol} \approx 0.0515 \ mol$.
Mass of $NaCl = 25 \ mL \times 0.116 = 2.9 \ g$.
Moles of $NaCl = \frac{2.9 \ g}{58.44 \ g/mol} \approx 0.0496 \ mol$.
Step $2$: Identify the limiting reagent.
Since the stoichiometry is $1:1$,$NaCl$ is the limiting reagent $(0.0496 \ mol < 0.0515 \ mol)$.
Step $3$: Calculate the mass of $AgCl$ formed.
Moles of $AgCl$ formed = Moles of $NaCl = 0.0496 \ mol$.
Mass of $AgCl = 0.0496 \ mol \times 143.32 \ g/mol \approx 7.1 \ g$.
Rounding to the nearest given option,the answer is $7 \ g$.

(B) The balanced chemical equation is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$
Molar mass of $CaCO_3 = 100 \ g/mol$
Molar mass of $HCl = 36.5 \ g/mol$
Molar mass of $CO_2 = 44 \ g/mol$
Moles of $CaCO_3 = \frac{20}{100} = 0.2 \ mol$
Moles of $HCl = \frac{20}{36.5} = 0.548 \ mol$
According to the reaction,$1 \ mol$ of $CaCO_3$ requires $2 \ mol$ of $HCl$.
So,$0.2 \ mol$ of $CaCO_3$ requires $0.2 \times 2 = 0.4 \ mol$ of $HCl$.
Since we have $0.548 \ mol$ of $HCl$,$CaCO_3$ is the limiting reagent.
Mass of $CO_2$ formed = $0.2 \ mol \times 44 \ g/mol = 8.8 \ g$.

(A) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is:
$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$
Calculate the moles of reactants:
Molar mass of $C_2H_6 = 30 \ g/mol$. Moles of $C_2H_6 = 15 \ g / 30 \ g/mol = 0.5 \ mol$.
Molar mass of $O_2 = 32 \ g/mol$. Moles of $O_2 = 112 \ g / 32 \ g/mol = 3.5 \ mol$.
Determine the limiting reagent:
For $0.5 \ mol$ of $C_2H_6$,required $O_2 = (7/2) \times 0.5 = 1.75 \ mol$.
Since we have $3.5 \ mol$ of $O_2$,$C_2H_6$ is the limiting reagent.
Calculate moles of products and remaining reactants:
$C_2H_6$ consumed = $0.5 \ mol$ (Remaining = $0 \ mol$).
$O_2$ consumed = $1.75 \ mol$ (Remaining = $3.5 - 1.75 = 1.75 \ mol$).
$CO_2$ produced = $(4/2) \times 0.5 = 1.0 \ mol$.
$H_2O$ produced = $(6/2) \times 0.5 = 1.5 \ mol$.
Total moles of gaseous substances = $n(O_2) + n(CO_2) + n(H_2O) = 1.75 + 1.0 + 1.5 = 4.25 \ mol$.

(D) Assume $100 \text{ moles}$ of the mixture. Then,$n_A = 50 \text{ mol}$,$n_B = 30 \text{ mol}$,and $n_C = 20 \text{ mol}$.
In reaction $1$: $A + 2B \rightarrow P_1$.
For $50 \text{ mol}$ of $A$,we need $100 \text{ mol}$ of $B$. Since we only have $30 \text{ mol}$ of $B$,$B$ is the limiting reagent.
$30 \text{ mol}$ of $B$ will react with $15 \text{ mol}$ of $A$ to produce $15 \text{ mol}$ of $P_1$.
Remaining $A = 50 - 15 = 35 \text{ mol}$.
In reaction $2$: $4P_1 + 3C \rightarrow P_2$.
We have $15 \text{ mol}$ of $P_1$ and $20 \text{ mol}$ of $C$.
For $15 \text{ mol}$ of $P_1$,the required amount of $C$ is $\frac{3}{4} \times 15 = 11.25 \text{ mol}$.
Since we have $20 \text{ mol}$ of $C$,$P_1$ is the limiting reagent in the second reaction.
$15 \text{ mol}$ of $P_1$ will react with $11.25 \text{ mol}$ of $C$ to produce $P_2$.
Remaining $C = 20 - 11.25 = 8.75 \text{ mol}$.
Remaining $A = 35 \text{ mol}$.
Since $P_1$ is completely consumed in the second reaction,$P_1$ will be completely exhausted.

(C) Step $1$: Calculate the moles of reactants.
Molar mass of $Cu \approx 63.5 \ g/mol$,Molar mass of $I_2 \approx 253.8 \ g/mol$.
Moles of $Cu = \frac{10}{63.5} \approx 0.157 \ mol$.
Moles of $I_2 = \frac{10}{253.8} \approx 0.0394 \ mol$.
Step $2$: Identify the limiting reagent.
According to the equation $2Cu + I_2 \longrightarrow 2CuI$,$1 \ mol$ of $I_2$ requires $2 \ mol$ of $Cu$.
For $0.0394 \ mol$ of $I_2$,we need $2 \times 0.0394 = 0.0788 \ mol$ of $Cu$.
Since we have $0.157 \ mol$ of $Cu$ (which is $> 0.0788 \ mol$),$I_2$ is the limiting reagent.
Step $3$: Calculate the yield of $CuI$.
Moles of $CuI$ produced $= 2 \times \text{moles of } I_2 = 2 \times 0.0394 = 0.0788 \ mol$.
Molar mass of $CuI = 63.5 + 126.9 = 190.4 \ g/mol$.
Mass of $CuI = 0.0788 \ mol \times 190.4 \ g/mol \approx 15 \ g$.
Thus,the correct option is $C$.

(B) The reaction is: $HCl + NaOH \rightarrow NaCl + H_2O$.
Number of moles of $HCl = 0.1 \ M \times 0.02 \ L = 0.002 \ mol$.
Number of moles of $NaOH = 0.1 \ M \times 0.03 \ L = 0.003 \ mol$.
Since $HCl$ is the limiting reagent,it will be completely consumed.
Number of moles of $NaOH$ remaining after neutralization $= 0.003 \ mol - 0.002 \ mol = 0.001 \ mol$.
Total final volume $= 20 \ mL + 30 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$.
Molarity of the final solution $= \frac{0.001 \ mol}{0.1 \ L} = 0.01 \ M$.

(D) The balanced chemical equation for the reaction is: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$.
According to the stoichiometry,$2 \ volumes$ of $H_2$ react with $1 \ volume$ of $O_2$.
Given initial volumes: $V(H_2) = 30 \ mL$ and $V(O_2) = 20 \ mL$.
Since $H_2$ is the limiting reagent,all $30 \ mL$ of $H_2$ will be consumed.
The volume of $O_2$ required to react with $30 \ mL$ of $H_2$ is $30 \ mL \times (1/2) = 15 \ mL$.
Therefore,the leftover volume of $O_2$ is $20 \ mL - 15 \ mL = 5 \ mL$ of $O_2$.

(A) Yeast and certain bacteria perform ethanol fermentation where glucose is broken down into ethanol and carbon dioxide.
The net chemical equation for the fermentation of glucose is:
$C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2$
From the balanced chemical equation,it is clear that $1 \text{ mole}$ of glucose produces $2 \text{ moles}$ of $CO_2$.

(C) By balancing the chemical equations:
$I. 3Fe_2O_3 + CO \longrightarrow 2Fe_3O_4 + CO_2 \uparrow$ (So,$P = Fe_3O_4$)
$II. Fe_3O_4 + 4CO \longrightarrow 3Fe + 4CO_2 \uparrow$ (So,$Q = Fe$)
$III. Fe_2O_3 + CO \longrightarrow 2FeO + CO_2 \uparrow$ (So,$R = FeO$)
Therefore,the correct option is $(c)$.

(B) Moles of $KI = 0.1 \times 0.250 = 0.025 \ mol$.
Moles of $KMnO_4 = 0.02 \times 0.250 = 0.005 \ mol$.
The balanced chemical equation is: $2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-$.
According to the stoichiometry,$2 \ moles$ of $KMnO_4$ react with $6 \ moles$ of $KI$ to produce $3 \ moles$ of $I_2$.
Here,$KMnO_4$ is the limiting reagent because $0.005 \ mol$ of $KMnO_4$ requires $0.015 \ mol$ of $KI$ (which is less than the $0.025 \ mol$ available).
Moles of $I_2$ formed $= \frac{3}{2} \times \text{moles of } KMnO_4 = \frac{3}{2} \times 0.005 = 0.0075 \ mol$.

(D) The balanced chemical equation for the reaction is: $2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 + I_2$.
Moles of $Cu^{2+} = M \times V(L) = 0.05 \times 0.1 = 5 \times 10^{-3} \ mol$.
Moles of $I^- = M \times V(L) = 0.1 \times 1 = 0.1 \ mol$.
Since $I^-$ is in excess,the reaction is limited by $Cu^{2+}$.
According to the stoichiometry,$2 \ mol$ of $Cu^{2+}$ produces $1 \ mol$ of $I_2$ and $1 \ mol$ of $Cu_2I_2$.
Therefore,$5 \times 10^{-3} \ mol$ of $Cu^{2+}$ will produce:
Moles of $I_2 = \frac{5 \times 10^{-3}}{2} = 2.5 \times 10^{-3} \ mol$.
Moles of $Cu_2I_2 = \frac{5 \times 10^{-3}}{2} = 2.5 \times 10^{-3} \ mol$.

(D) The given unbalanced reaction is: $aKNO_3 + 5C_{12}H_{22}O_{11} \rightarrow bN_2 + cCO_2 + dH_2O + eK_2CO_3$.
To balance the equation,we equate the number of atoms of each element on both sides:
For $K$: $a = 2e$
For $N$: $a = 2b$
For $C$: $5 \times 12 = c + e \Rightarrow 60 = c + e$
For $H$: $5 \times 22 = 2d$ $\Rightarrow 110 = 2d$ $\Rightarrow d = 55$
For $O$: $3a + 5 \times 11 = 2c + d + 3e$ $\Rightarrow 3a + 55 = 2c + 55 + 3e$ $\Rightarrow 3a = 2c + 3e$
Setting $e = 24$,we get $a = 48$.
Since $a = 2b$,$b = 24$.
Since $c + e = 60$,$c = 60 - 24 = 36$.
Thus,the balanced equation is: $48KNO_3 + 5C_{12}H_{22}O_{11} \rightarrow 24N_2 + 36CO_2 + 55H_2O + 24K_2CO_3$.
The values are $A=48, B=24, C=36, D=55, E=24$.