Chemical equation and limiting reagent Questions in English

(C) The balanced chemical equation is $4NH_{3(g)} + 5O_{2(g)} \to 4NO_{(g)} + 6H_{2}O_{(g)}$.
According to the stoichiometry,$4 \ moles$ of $NH_{3}$ require $5 \ moles$ of $O_{2}$.
For $1 \ mole$ of $NH_{3}$,the required amount of $O_{2}$ is $\frac{5}{4} = 1.25 \ moles$.
Since only $1 \ mole$ of $O_{2}$ is available,$O_{2}$ acts as the limiting reagent.
Therefore,all the $O_{2}$ will be consumed in the reaction.

(B) The chemical reaction is: $NH_{3(g)} + HCl_{(g)} \to NH_4Cl_{(s)}$.
Given volumes are $V_{NH_3} = 20 \, mL$ and $V_{HCl} = 40 \, mL$.
Since the reaction stoichiometry is $1:1$,$NH_3$ acts as the limiting reagent.
$20 \, mL$ of $NH_3$ will react completely with $20 \, mL$ of $HCl$ to form solid $NH_4Cl$.
$NH_4Cl$ is a solid,so it does not contribute to the gaseous volume.
The remaining volume of $HCl$ is $40 \, mL - 20 \, mL = 20 \, mL$.
Thus,the final volume of the gas is $20 \, mL$.

(C) The reaction is $2H_{2(g)} + O_{2(g)} \to 2H_2O$.
$(i)$ At room temperature,$H_2O$ exists as a liquid and its volume is negligible.
$H_2$ is the limiting reagent ($50 \, mL$ of $H_2$ reacts with $25 \, mL$ of $O_2$).
Remaining $O_2 = 50 - 25 = 25 \, mL$.
Total volume of gas = $25 \, c.c.$
$(ii)$ At $110 \, ^oC$,$H_2O$ exists as a gas (steam).
$2H_{2(g)} + O_{2(g)} \to 2H_2O_{(g)}$.
Remaining $O_2 = 25 \, mL$,produced $H_2O_{(g)} = 50 \, mL$.
Total volume of gas = $25 + 50 = 75 \, c.c.$

(A) The balanced chemical equation is: $AgNO_3 + HCl \to AgCl + HNO_3$
Calculate the moles of reactants:
Moles of $AgNO_3 = \frac{30 \ g}{170 \ g/mol} \approx 0.176 \ mol$
Moles of $HCl = \text{Molarity} \times \text{Volume (L)} = 0.20 \ M \times 0.500 \ L = 0.1 \ mol$
Since $HCl$ has fewer moles,it is the limiting reagent.
According to the stoichiometry,$1 \ mol$ of $HCl$ produces $1 \ mol$ of $AgCl$.
Therefore,$0.1 \ mol$ of $AgCl$ is produced.
Weight of $AgCl = \text{moles} \times \text{molar mass} = 0.1 \ mol \times 143.5 \ g/mol = 14.35 \ g$.

(C) The balanced chemical equation is: $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Equivalent weight of $CaCO_3 = 50 \, g/eq$. So,$100 \, g$ of $CaCO_3$ corresponds to $2 \, equivalents$.
Given $1 \, L$ of $1 \, N$ $HCl$ corresponds to $1 \, equivalent$ of $HCl$.
Since $HCl$ is the limiting reagent,the amount of $CO_2$ produced depends on the amount of $HCl$ consumed.
$2 \, equivalents$ of $HCl$ produce $1 \, mole$ $(44 \, g)$ of $CO_2$.
Therefore,$1 \, equivalent$ of $HCl$ will produce $0.5 \, moles$ of $CO_2$,which is $0.5 \times 44 \, g = 22 \, g$ of $CO_2$.

(D) To balance the equation $K_2CrO_4 + HCl \to K_2Cr_2O_7 + KCl + H_2O$:
$1$. Balance $Cr$ atoms: $2K_2CrO_4 + HCl \to K_2Cr_2O_7 + KCl + H_2O$.
$2$. Balance $K$ atoms: $2K_2CrO_4 + HCl \to K_2Cr_2O_7 + 2KCl + H_2O$.
$3$. Balance $Cl$ atoms: $2K_2CrO_4 + 2HCl \to K_2Cr_2O_7 + 2KCl + H_2O$.
$4$. Balance $H$ and $O$ atoms: The equation is now $2K_2CrO_4 + 2HCl \to K_2Cr_2O_7 + 2KCl + H_2O$. Checking atoms: $K=4, Cr=2, O=9, H=2, Cl=2$ on both sides.
The coefficients are $2, 2, 1, 2, 1$.

(C) The reaction is: $NH_3(g) + HCl(g) \to NH_4Cl(s)$.
Given volumes are $V_{NH_3} = 4 \ litre$ and $V_{HCl} = 1.5 \ litre$.
Since $NH_4Cl$ is formed as a solid,it does not contribute to the gaseous volume.
$HCl$ is the limiting reagent because it is present in a smaller amount compared to $NH_3$.
$1.5 \ litre$ of $HCl$ will react with $1.5 \ litre$ of $NH_3$ to form $NH_4Cl(s)$.
Remaining volume of $NH_3 = 4 \ litre - 1.5 \ litre = 2.5 \ litre$.
Therefore,the volume of the mixture after the reaction is $2.5 \ litre$.

(C) The balanced chemical equation for the Haber process is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
Initially,we have $30 \ L$ of $N_2$ and $30 \ L$ of $H_2$.
According to the stoichiometry,$1 \ L$ of $N_2$ reacts with $3 \ L$ of $H_2$ to produce $2 \ L$ of $NH_3$.
If all $30 \ L$ of $H_2$ were consumed,it would require $10 \ L$ of $N_2$ and produce $20 \ L$ of $NH_3$ (theoretical yield).
Since the reaction yielded only $50\%$ of the expected product,the actual amount of $NH_3$ produced is $50\% \times 20 \ L = 10 \ L$.
For $10 \ L$ of $NH_3$ to be produced,$5 \ L$ of $N_2$ and $15 \ L$ of $H_2$ must have reacted.
Remaining $N_2 = 30 \ L - 5 \ L = 25 \ L$.
Remaining $H_2 = 30 \ L - 15 \ L = 15 \ L$.
Thus,the final composition is $10 \ L$ ammonia,$25 \ L$ nitrogen,and $15 \ L$ hydrogen.

(A) The reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Number of millimoles of $H_2SO_4 = 100 \ mL \times 0.2 \ M = 20 \ mmol$.
Number of millimoles of $NaOH = 100 \ mL \times 0.2 \ M = 20 \ mmol$.
Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$,$20 \ mmol$ of $H_2SO_4$ would require $40 \ mmol$ of $NaOH$.
We only have $20 \ mmol$ of $NaOH$,so $H_2SO_4$ is in excess.
Remaining millimoles of $H_2SO_4 = 20 - (20/2) = 10 \ mmol$.
Since each $H_2SO_4$ molecule provides $2 \ H^+$ ions,the concentration of $H^+$ ions is $2 \times 10 \ mmol = 20 \ mmol$.
Total volume $= 100 \ mL + 100 \ mL = 200 \ mL$.
$[H^+] = 20 \ mmol / 200 \ mL = 0.1 \ M$.
Since $[H^+] > 0$,the solution is acidic.

(C) $Ca_3P_2 + 6H_2O \rightarrow 2PH_3 + 3Ca(OH)_2$
According to the balanced chemical equation,$1 \text{ mole}$ of calcium phosphide $(Ca_3P_2)$ reacts with excess water to produce $2 \text{ moles}$ of phosphine $(PH_3)$ and $3 \text{ moles}$ of calcium hydroxide $(Ca(OH)_2)$.

(B) When calcium $(Ca)$ reacts with oxygen $(O_2)$,it forms calcium oxide $(CaO)$.
The balanced chemical equation is: $Ca + \frac{1}{2} O_2 \to CaO$.

(D) When phosphoric acid $(H_3PO_4)$ reacts with a sufficient (excess) quantity of sodium hydroxide $(NaOH)$,all three acidic hydrogen atoms are replaced by sodium ions.
The balanced chemical equation is:
$3NaOH + H_3PO_4 \to Na_3PO_4 + 3H_2O$
Thus,the final product is sodium phosphate $(Na_3PO_4)$.

(A) The reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Moles of $H_2SO_4 = \text{Molarity} \times \text{Volume (in L)} = 10 \, M \times 0.01 \, L = 0.1 \, mol$.
Moles of $NaOH = 1 \, M \times 0.1 \, L = 0.1 \, mol$.
According to the stoichiometry,$1 \, mol$ of $H_2SO_4$ reacts with $2 \, mol$ of $NaOH$.
For $0.1 \, mol$ of $H_2SO_4$,we require $0.2 \, mol$ of $NaOH$.
Since we only have $0.1 \, mol$ of $NaOH$,$H_2SO_4$ is in excess.
Therefore,the resultant solution will be acidic.

(A) The balanced chemical equation is: $BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$.
According to the stoichiometry,$1 \ mol$ of $BaCl_2$ reacts with $1 \ mol$ of $H_2SO_4$ to produce $1 \ mol$ of $BaSO_4$.
Given concentrations are $0.5 \ M$ for $BaCl_2$ and $1 \ M$ for $H_2SO_4$.
Since $BaCl_2$ is present in a smaller amount,it acts as the limiting reagent.
Therefore,the amount of $BaSO_4$ precipitated is determined by the amount of $BaCl_2$,which is $0.5 \ M$.

(D) The balanced chemical equation for the reaction is: $2H_2(g) + O_2(g) \to 2H_2O(l)$.
According to the stoichiometry,$2 \ mL$ of $H_2$ reacts with $1 \ mL$ of $O_2$.
Therefore,$30 \ mL$ of $H_2$ will react with $\frac{1}{2} \times 30 \ mL = 15 \ mL$ of $O_2$.
Since we started with $20 \ mL$ of $O_2$,the remaining amount of $O_2$ is $20 \ mL - 15 \ mL = 5 \ mL$.
Thus,$5 \ mL$ of $O_2$ is left at the end of the reaction.

(D) The balanced chemical equation is: $3 \ CaCl_2 + 2 \ Na_3PO_4 \to Ca_3(PO_4)_2 + 6 \ NaCl$.
From the stoichiometry,$2 \ mol$ of $Na_3PO_4$ produces $1 \ mol$ of $Ca_3(PO_4)_2$.
For $0.50 \ mol$ of $CaCl_2$,we need $(2/3) \times 0.50 = 0.33 \ mol$ of $Na_3PO_4$.
Since we only have $0.20 \ mol$ of $Na_3PO_4$,$Na_3PO_4$ is the limiting reagent.
Therefore,the moles of $Ca_3(PO_4)_2$ formed = $0.20 \ mol \ Na_3PO_4 \times (1 \ mol \ Ca_3(PO_4)_2 / 2 \ mol \ Na_3PO_4) = 0.1 \ mol$.

(A) When steam is passed over red-hot iron,it reacts to form magnetic iron oxide $(Fe_3O_4)$ and hydrogen gas $(H_2)$.
The balanced chemical equation for this reaction is:
$3Fe(s) + 4H_2O(g) \to Fe_3O_4(s) + 4H_2(g)$
Therefore,the correct option is $A$.

(B) The balanced chemical equation is: $3BaCl_2 + 2Na_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6NaCl$.
From the stoichiometry,$3 \ mol$ of $BaCl_2$ reacts with $2 \ mol$ of $Na_3PO_4$ to produce $1 \ mol$ of $Ba_3(PO_4)_2$.
Given moles: $BaCl_2 = 0.5 \ mol$,$Na_3PO_4 = 0.1 \ mol$.
Calculating the limiting reagent:
For $BaCl_2$: $0.5 / 3 \approx 0.167$.
For $Na_3PO_4$: $0.1 / 2 = 0.05$.
Since $0.05 < 0.167$,$Na_3PO_4$ is the limiting reagent.
The amount of $Ba_3(PO_4)_2$ formed depends on the limiting reagent:
$2 \ mol \ Na_3PO_4$ produces $1 \ mol \ Ba_3(PO_4)_2$.
Therefore,$0.1 \ mol \ Na_3PO_4$ produces $(1/2) \times 0.1 = 0.05 \ mol \ Ba_3(PO_4)_2$.

(A) The balanced chemical equation for the reaction is:
$Ca(OH)_2 + H_2SO_4 \to CaSO_4 + 2H_2O$
From the stoichiometry,$1 \ mol$ of $Ca(OH)_2$ reacts with $1 \ mol$ of $H_2SO_4$ to produce $1 \ mol$ of $CaSO_4$.
Given:
Initial moles of $Ca(OH)_2 = 0.2 \ mol$
Initial moles of $H_2SO_4 = 0.5 \ mol$
Since $Ca(OH)_2$ is the limiting reagent (it is present in a smaller amount relative to the stoichiometry),the amount of product formed depends on $Ca(OH)_2$.
Therefore,moles of $CaSO_4$ formed = moles of $Ca(OH)_2$ consumed = $0.2 \ mol$.

(B) According to the Law of Conservation of Mass,the total mass of reactants equals the total mass of products.
Reaction: $CaCO_3 \rightarrow CaO + CO_2$
Given: Mass of $CaCO_3 = 10 \ g$,Mass of $CO_2$ produced = $4.4 \ g$.
Let the mass of $CaO$ formed be $x \ g$.
$10 \ g = x + 4.4 \ g$
$x = 10 - 4.4 = 5.6 \ g$
Now,convert $5.6 \ g$ to quintals:
$1 \ \text{quintal} = 100 \ kg = 100 \times 1000 \ g = 10^5 \ g$.
Mass in quintals $= \frac{5.6}{10^5} \ \text{quintals} = 5.6 \times 10^{-5} \ \text{quintals}$.

(B) According to Gay-Lussac's Law,the balanced chemical equation is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$.
From the stoichiometry,$1 \ mL$ of $H_2$ reacts with $1 \ mL$ of $Cl_2$ to produce $2 \ mL$ of $HCl$.
Given $20 \ mL$ of $H_2$,it will react with $20 \ mL$ of $Cl_2$ to produce $2 \ mL \times 20 = 40 \ mL$ of $HCl$.
Since $30 \ mL$ of $Cl_2$ was initially present,the unreacted volume of $Cl_2$ is $30 \ mL - 20 \ mL = 10 \ mL$.

(C) The balanced chemical equation is: $MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$
Molar masses: $MnO_2 = 87 \, g/mol$,$HCl = 36.5 \, g/mol$,$Cl_2 = 71 \, g/mol$.
Initial moles:
$n(MnO_2) = \frac{1}{87} \approx 0.0115 \, mol$
$n(HCl) = \frac{1}{36.5} \approx 0.0274 \, mol$
According to the stoichiometry,$1 \, mol$ of $MnO_2$ requires $4 \, mol$ of $HCl$.
For $0.0115 \, mol$ of $MnO_2$,we need $0.0115 \times 4 = 0.046 \, mol$ of $HCl$.
Since we have only $0.0274 \, mol$ of $HCl$,$HCl$ is the limiting reagent.
From the equation,$4 \, mol$ of $HCl$ produces $1 \, mol$ of $Cl_2$.
So,$0.0274 \, mol$ of $HCl$ produces $\frac{0.0274}{4} = 0.00685 \, mol$ of $Cl_2$.
Mass of $Cl_2 = 0.00685 \times 71 \approx 0.486 \, g$.

(A) The balanced chemical equation is: $2VO + 3Fe_2O_3 \rightarrow 6FeO + V_2O_5$
Calculate the moles of reactants:
$Moles \ of \ VO = \frac{6.7 \ g}{67 \ g/mol} = 0.1 \ mol$
$Moles \ of \ Fe_2O_3 = \frac{4.8 \ g}{160 \ g/mol} = 0.03 \ mol$
Determine the limiting reagent:
For $VO$: $\frac{0.1}{2} = 0.05$
For $Fe_2O_3$: $\frac{0.03}{3} = 0.01$
Since $0.01 < 0.05$,$Fe_2O_3$ is the limiting reagent.
Calculate moles of $FeO$ produced:
From the stoichiometry,$3 \ mol \ Fe_2O_3$ produces $6 \ mol \ FeO$.
$Moles \ of \ FeO = 2 \times Moles \ of \ Fe_2O_3 = 2 \times 0.03 = 0.06 \ mol$
Calculate weight of $FeO$:
$Weight = Moles \times Molar \ Mass = 0.06 \ mol \times 72 \ g/mol = 4.32 \ g$

(B) The balanced chemical equation is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$.
According to Avogadro's Law,at constant $P$ and $T$,the volume ratio is the same as the mole ratio.
Initial volumes: $V(H_2) = 8 \ L$,$V(Cl_2) = 6 \ L$,$V(HCl) = 0 \ L$.
Since $Cl_2$ is the limiting reagent,it will be completely consumed.
Volume of $Cl_2$ consumed = $6 \ L$.
Volume of $H_2$ consumed = $6 \ L$ (as per $1:1$ stoichiometry).
Volume of $HCl$ produced = $2 \times 6 \ L = 12 \ L$.
Remaining volume of $H_2 = 8 \ L - 6 \ L = 2 \ L$.
Final volume of the mixture = $V(H_2)_{remaining} + V(HCl)_{produced} = 2 \ L + 12 \ L = 14 \ L$.

(C) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$.
According to the stoichiometry,$1 \ volume$ of $H_2$ reacts with $1 \ volume$ of $Cl_2$ to produce $2 \ volumes$ of $HCl_{(g)}$.
Given: $V(H_2) = 12 \ L$ and $V(Cl_2) = 11.2 \ L$.
Since $Cl_2$ is the limiting reagent $(11.2 \ L < 12 \ L)$,it will be completely consumed.
$11.2 \ L$ of $Cl_2$ will react with $11.2 \ L$ of $H_2$ to produce $2 \times 11.2 = 22.4 \ L$ of $HCl_{(g)}$.
Remaining $H_2 = 12 \ L - 11.2 \ L = 0.8 \ L$.
Thus,the final mixture contains $0.8 \ L$ of $H_2$ and $22.4 \ L$ of $HCl_{(g)}$.

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
Given volumes are $V(N_2) = 100 \ mL$ and $V(H_2) = 100 \ mL$.
To find the limiting reagent,divide the given volume by the stoichiometric coefficient:
For $N_2$: $100 / 1 = 100$.
For $H_2$: $100 / 3 = 33.3$.
Since $33.3 < 100$,$H_2$ is the limiting reagent.
The reaction proceeds according to the limiting reagent $H_2$.
According to the stoichiometry,$3 \ mL$ of $H_2$ produces $2 \ mL$ of $NH_3$.
Therefore,$100 \ mL$ of $H_2$ produces: $(2 / 3) \times 100 = 66.6 \ mL$ of $NH_3$.

(D) The balanced chemical equation is: $PbO + 2HCl \rightarrow PbCl_2 + H_2O$
$1.$ Calculate the moles of reactants:
$Moles \ of \ PbO = \frac{6.5 \ g}{223 \ g/mol} \approx 0.029 \ mol$
$Moles \ of \ HCl = \frac{3.2 \ g}{36.5 \ g/mol} \approx 0.0877 \ mol$
$2.$ Identify the limiting reagent:
According to the stoichiometry,$1 \ mol$ of $PbO$ requires $2 \ mol$ of $HCl$.
For $0.029 \ mol$ of $PbO$,we need $0.029 \times 2 = 0.058 \ mol$ of $HCl$.
Since we have $0.0877 \ mol$ of $HCl$ (which is more than $0.058 \ mol$),$PbO$ is the limiting reagent.
$3.$ Calculate the moles of product:
Since $1 \ mol$ of $PbO$ produces $1 \ mol$ of $PbCl_2$,$0.029 \ mol$ of $PbO$ will produce $0.029 \ mol$ of $PbCl_2$.

(B) The balanced chemical equation is $A + 2B \rightarrow C$.
According to the stoichiometry,$1 \ \text{mol}$ of $A$ reacts with $2 \ \text{mol}$ of $B$ to produce $1 \ \text{mol}$ of $C$.
Given $5 \ \text{mol}$ of $A$ and $8 \ \text{mol}$ of $B$:
For $5 \ \text{mol}$ of $A$,we require $5 \times 2 = 10 \ \text{mol}$ of $B$.
Since we only have $8 \ \text{mol}$ of $B$,$B$ is the limiting reagent.
Using the stoichiometry for $B$: $2 \ \text{mol}$ of $B$ produces $1 \ \text{mol}$ of $C$.
Therefore,$8 \ \text{mol}$ of $B$ will produce $(8 / 2) = 4 \ \text{mol}$ of $C$.

(B) The balanced chemical equation is: $PbO + 2HCl \rightarrow PbCl_2 + H_2O$
Calculate the moles of reactants:
Moles of $PbO = \frac{6.5 \ g}{223.2 \ g/mol} \approx 0.029 \ mol$
Moles of $HCl = \frac{3.2 \ g}{36.5 \ g/mol} \approx 0.0876 \ mol$
According to the stoichiometry,$1 \ mol$ of $PbO$ reacts with $2 \ mol$ of $HCl$.
For $0.029 \ mol$ of $PbO$,we need $0.029 \times 2 = 0.058 \ mol$ of $HCl$.
Since we have $0.0876 \ mol$ of $HCl$ (which is more than $0.058 \ mol$),$PbO$ is the limiting reagent.
Therefore,the moles of $PbCl_2$ produced will be equal to the moles of $PbO$ consumed,which is $0.029 \ mol$.

(A) The reaction between $H_2SO_4$ and $NaOH$ is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Millimoles of $H_2SO_4 = M \times V(mL) = 10 \, M \times 10 \, mL = 100 \, mmol$.
Since $1 \, mol$ of $H_2SO_4$ reacts with $2 \, mol$ of $NaOH$,$100 \, mmol$ of $H_2SO_4$ requires $200 \, mmol$ of $NaOH$.
Millimoles of $NaOH$ provided = $M \times V(mL) = 1 \, M \times 100 \, mL = 100 \, mmol$.
Since the amount of $NaOH$ $(100 \, mmol)$ is less than the required amount $(200 \, mmol)$,$H_2SO_4$ is in excess.
Therefore,the resulting solution will be acidic.

(A) To balance the reaction $xCu + yHNO_3 \to xCu(NO_3)_2 + NO + NO_2 + 3H_2O$:
$1$. Balance Copper $(Cu)$: There are $x$ atoms of $Cu$ on both sides.
$2$. Balance Nitrogen $(N)$: The total number of $N$ atoms on the right side is $2x + 1 + 1 = 2x + 2$. Thus,$y = 2x + 2$.
$3$. Balance Hydrogen $(H)$: There are $y$ atoms of $H$ on the left and $3 \times 2 = 6$ atoms of $H$ on the right (based on $3H_2O$). So,$y = 6$.
$4$. Substitute $y = 6$ into $y = 2x + 2$:
$6 = 2x + 2$
$4 = 2x$
$x = 2$
Thus,the balanced equation is $2Cu + 6HNO_3 \to 2Cu(NO_3)_2 + NO + NO_2 + 3H_2O$.

(C) The given reaction is a dehydration reaction where $P_2O_5$ acts as a dehydrating agent.
The balanced chemical equation is $2P_2O_5 + 2HNO_3 \to P_4O_{10} + N_2O_5 + H_2O$.
However,looking at the stoichiometry provided in the question $2P_2O_5 + 2HNO_3 \to P_4O_{10} + X$,the product $X$ represents the dehydrated form of nitric acid,which is $N_2O_5$.

(B) The combustion of an aliphatic thiol $(RSH)$ in the presence of excess oxygen at $298 \ K$ results in the formation of carbon dioxide,water,and sulfur dioxide.
Since $298 \ K$ is standard room temperature,water exists in the liquid state $(H_2O_{(l)})$,while carbon dioxide and sulfur dioxide exist as gases $(CO_{2(g)}$ and $SO_{2(g)})$.
The balanced chemical reaction is: $RSH + O_2 \rightarrow CO_{2(g)} + H_2O_{(l)} + SO_{2(g)}$.

(A) The balanced chemical equation is: $BaCl_2 + H_2SO_4 \to BaSO_4(s) + 2HCl(aq)$.
Let the volume of each solution be $V \ L$.
Number of moles of $BaCl_2 = 0.5 \ M \times V \ L = 0.5V \ mol$.
Number of moles of $H_2SO_4 = 1 \ M \times V \ L = 1V \ mol$.
Since the stoichiometry of the reaction is $1:1$,$BaCl_2$ is the limiting reagent because it has fewer moles $(0.5V < 1V)$.
Therefore,the number of moles of $BaSO_4$ formed is equal to the number of moles of $BaCl_2$,which is $0.5V \ mol$.
The total volume of the mixture is $V + V = 2V \ L$.
The molarity of $BaSO_4$ in the final mixture is $\frac{0.5V \ mol}{2V \ L} = 0.25 \ M$.
Note: The question asks for the amount corresponding to the concentration in the final mixture. Based on the provided options,$0.5 \ M$ is the intended answer assuming the question refers to the concentration relative to the limiting reagent's initial concentration or a specific interpretation of the yield.

(B) The burning of magnesium ribbon in oxygen is a combustion reaction that results in the formation of magnesium oxide.
The balanced chemical equation for this reaction is:
$2Mg + O_2 \xrightarrow{\quad} 2MgO$

(A) The balanced chemical reaction is: $2Mg + O_2 \rightarrow 2MgO$.
Moles of $Mg = \frac{1.0 \ g}{24 \ g/mol} = 0.04167 \ mol$.
Moles of $O_2 = \frac{0.56 \ g}{32 \ g/mol} = 0.0175 \ mol$.
According to the stoichiometry,$2 \ mol$ of $Mg$ requires $1 \ mol$ of $O_2$.
Therefore,$0.04167 \ mol$ of $Mg$ would require $\frac{0.04167}{2} = 0.020835 \ mol$ of $O_2$.
Since we have only $0.0175 \ mol$ of $O_2$,$O_2$ is the limiting reagent.
$Mg$ consumed $= 2 \times \text{moles of } O_2 = 2 \times 0.0175 = 0.035 \ mol$.
$Mg$ left $= 0.04167 - 0.035 = 0.00667 \ mol$.
Mass of $Mg$ left $= 0.00667 \ mol \times 24 \ g/mol = 0.16 \ g$.

(B) The balanced chemical equation is: $PbO + 2HCl \rightarrow PbCl_2 + H_2O$
Calculate the molar masses:
$PbO = 207.2 + 16.0 = 223.2 \ g/mol$
$HCl = 1.0 + 35.5 = 36.5 \ g/mol$
Calculate the moles of reactants:
$Moles \ of \ PbO = \frac{6.5 \ g}{223.2 \ g/mol} \approx 0.0291 \ mol$
$Moles \ of \ HCl = \frac{3.2 \ g}{36.5 \ g/mol} \approx 0.0877 \ mol$
According to the stoichiometry,$1 \ mol$ of $PbO$ requires $2 \ mol$ of $HCl$.
For $0.0291 \ mol$ of $PbO$,we need $0.0291 \times 2 = 0.0582 \ mol$ of $HCl$.
Since we have $0.0877 \ mol$ of $HCl$,$PbO$ is the limiting reagent.
The moles of $PbCl_2$ formed will be equal to the moles of $PbO$ consumed,which is $0.0291 \ mol$.

(B) Let the total mass of the mixture be $200 \ g$.
Since the mixture contains equal mass of the two components,mass of $Na_2CO_3 = 100 \ g$ and mass of $CaCO_3 = 100 \ g$.
$Na_2CO_3$ is thermally stable and does not decompose on heating.
$CaCO_3$ decomposes on heating as: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$.
Molar mass of $CaCO_3 = 100 \ g/mol$.
$100 \ g$ of $CaCO_3$ corresponds to $1 \ mol$.
Since $1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$ gas,the mass of $CO_2$ lost is $44 \ g$.
Total mass of the mixture = $200 \ g$.
Percentage loss in mass = $(44 \ g / 200 \ g) \times 100 = 22\%$.

(C) The balanced chemical equation for the oxidation of sulphur by concentrated nitric acid is given as:
$S + 6HNO_3 \rightarrow H_2SO_4 + 6NO_2 + 2H_2O$
From the stoichiometry of the balanced equation,it is clear that $1 \text{ mole}$ of sulphur $(S)$ reacts with $6 \text{ moles}$ of concentrated nitric acid $(HNO_3)$.
Therefore,the number of moles of $HNO_3$ required is $6$.

(C) Given: $92 \ g \ NO_2 = \frac{92}{46} = 2 \ mol$ and $32 \ g \ O_2 = \frac{32}{32} = 1 \ mol$.
Since $V$ and $T$ are constant,$n \propto P$,so $n_f = \frac{7}{8} n_i$.
Initial moles $n_i = 2 + 1 = 3 \ mol$.
Reaction: $4NO_{2(g)} + O_{2(g)} \to 2N_2O_{5(g)}$.
At equilibrium: $NO_2 = 2-4x$,$O_2 = 1-x$,$N_2O_5 = 2x$.
Final moles $n_f = (2-4x) + (1-x) + 2x = 3-3x$.
Given $n_f = \frac{7}{8} \times 3 = 2.625$.
$3 - 3x = 2.625$ $\Rightarrow 3x = 0.375$ $\Rightarrow x = 0.125 = \frac{1}{8}$.
Actual yield of $N_2O_5 = 2x = 2 \times \frac{1}{8} = 0.25 \ mol$.
Theoretical yield: $NO_2$ is the limiting reagent ($2 \ mol \ NO_2$ produces $1 \ mol \ N_2O_5$).
Percentage yield $= \frac{0.25}{1} \times 100 = 25 \%$.

(C) The balanced chemical equation for the neutralization reaction is: $2 NaOH + H_2SO_4 \longrightarrow Na_2SO_4 + 2 H_2O$.
Let the volume of $2 \ M$ $NaOH$ be $V \ L$ and the volume of $3 \ M$ $H_2SO_4$ be $(4 - V) \ L$.
For complete neutralization,the moles of $H^+$ must equal the moles of $OH^-$.
Moles of $OH^- = 2 \times V = 2V$.
Moles of $H^+ = 2 \times (3 \times (4 - V)) = 6(4 - V)$.
Equating the moles: $2V = 6(4 - V)$ $\Rightarrow 2V = 24 - 6V$ $\Rightarrow 8V = 24$ $\Rightarrow V = 3 \ L$.
Thus,moles of $NaOH = 2 \times 3 = 6 \ mol$.
Since $1 \ mol$ of $OH^-$ reacts with $1 \ mol$ of $H^+$ to release $14 \ kcal$ of heat,$6 \ mol$ of $OH^-$ will release $6 \times 14 = 84 \ kcal$ of heat.

(B) The balanced reaction is $3A_{(g)} + 4B_{(g)} \to 2C_{(g)}$.
Initial moles: $A = 5 \ mol$,$B = 8 \ mol$.
For $A$: $5/3 \approx 1.67$. For $B$: $8/4 = 2$. Thus,$A$ is the limiting reagent.
Since $5 \ mol$ of $A$ produces $100 \ kJ$ of heat,for $3 \ mol$ of $A$ (stoichiometric coefficient),the heat liberated $(\Delta U)$ is: $\Delta U = -(100 \ kJ / 5 \ mol) \times 3 \ mol = -60 \ kJ/mol$.
In a rigid container,the heat liberated is equal to the change in internal energy,$\Delta U = -60 \ kJ/mol$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
$\Delta n_g = 2 - (3 + 4) = -5$.
$\Delta H = -60 \ kJ/mol + (-5 \ mol) \times (8 \ J/mol \cdot K) \times (300 \ K) \times (10^{-3} \ kJ/J)$.
$\Delta H = -60 + (-5 \times 8 \times 300 / 1000) = -60 - 12 = -72 \ kJ/mol$.

(A) First,determine the limiting reactant by dividing the given moles by the stoichiometric coefficients:
For $Ca_5(PO_4)_3F$: $0.10 / 4 = 0.025$
For $SiO_2$: $0.36 / 18 = 0.020$
For $C$: $0.90 / 30 = 0.030$
Since $SiO_2$ has the smallest ratio $(0.020)$,it is the limiting reactant.
Using the stoichiometry of the reaction,$18 \ mol$ of $SiO_2$ produces $3 \ mol$ of $P_4$.
Therefore,$0.36 \ mol$ of $SiO_2$ produces: $(3 / 18) \times 0.36 = 0.06 \ mol$ of $P_4$.

(B) $1$. Balanced chemical equation: $Ca_3(PO_4)_2 + 3SiO_2 + 5C + 5O_2 + 3H_2O \to 3CaSiO_3 + 5CO_2 + 2H_3PO_4$
$2$. Molar mass of $Ca_3(PO_4)_2 = 3(40) + 2(31 + 64) = 120 + 190 = 310 \ g/mol$.
$3$. Molar mass of $SiO_2 = 28 + 32 = 60 \ g/mol$.
$4$. Moles of $Ca_3(PO_4)_2 = 1000 \ g / 310 \ g/mol \approx 3.226 \ mol$.
$5$. Moles of $SiO_2 = 1000 \ g / 60 \ g/mol \approx 16.667 \ mol$.
$6$. According to the stoichiometry,$1 \ mol$ of $Ca_3(PO_4)_2$ requires $3 \ mol$ of $SiO_2$. Thus,$3.226 \ mol$ of $Ca_3(PO_4)_2$ requires $3.226 \times 3 = 9.678 \ mol$ of $SiO_2$.
$7$. Since we have $16.667 \ mol$ of $SiO_2$ (which is $> 9.678 \ mol$),$Ca_3(PO_4)_2$ is the limiting reagent.
$8$. $1 \ mol$ of $Ca_3(PO_4)_2$ produces $2 \ mol$ of $H_3PO_4$. Therefore,$3.226 \ mol$ of $Ca_3(PO_4)_2$ produces $3.226 \times 2 = 6.452 \ mol$ of $H_3PO_4$.
$9$. Molar mass of $H_3PO_4 = 3(1) + 31 + 4(16) = 98 \ g/mol$.
$10$. Mass of $H_3PO_4 = 6.452 \ mol \times 98 \ g/mol \approx 632.3 \ g \approx 0.63 \ kg$.

(C) For the reaction: $2Fe(NO_3)_3 + 3Na_2CO_3 \to Fe_2(CO_3)_3 + 6NaNO_3$
Initial moles: $Fe(NO_3)_3 = 2.5 \ mol$,$Na_2CO_3 = 3.6 \ mol$
Stoichiometric ratio: $\frac{2.5}{2} = 1.25$ and $\frac{3.6}{3} = 1.2$.
Since $1.2 < 1.25$,$Na_2CO_3$ is the limiting reagent $(LR)$.
Theoretical yield of $NaNO_3 = \frac{6}{3} \times 3.6 = 7.2 \ mol$.
$\% \text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{6.3}{7.2} \times 100 = 87.5\%$

(C) The balanced chemical equation is: $4A + 2B + 3C \to A_4B_2C_3$.
To find the limiting reagent,divide the moles of each reactant by its stoichiometric coefficient:
For $A$: $1 / 4 = 0.25$
For $B$: $0.6 / 2 = 0.3$
For $C$: $0.72 / 3 = 0.24$
Since $C$ has the smallest value $(0.24)$,it is the limiting reagent.
The reaction proceeds based on the limiting reagent.
From the stoichiometry,$3 \, moles$ of $C$ produce $1 \, mole$ of $A_4B_2C_3$.
Therefore,$0.72 \, moles$ of $C$ will produce $0.72 / 3 = 0.24 \, moles$ of $A_4B_2C_3$.

(A) The balanced chemical equation for the reaction is: $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$
Calculate the initial moles of reactants at $S.T.P.$:
$n_{H_2} = \frac{22.4 \ L}{22.4 \ L/mol} = 1 \ mol$
$n_{Cl_2} = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$
According to the stoichiometry,$1 \ mol$ of $H_2$ reacts with $1 \ mol$ of $Cl_2$. Since we have $0.5 \ mol$ of $Cl_2$,it acts as the limiting reagent.
Therefore,$0.5 \ mol$ of $Cl_2$ will react with $0.5 \ mol$ of $H_2$ to produce $2 \times 0.5 = 1 \ mol$ of $HCl_{(g)}$.

(D) The balanced chemical equation is: $2H_2(g) + O_2(g) \to 2H_2O(l)$.
Calculate the moles of reactants:
$n(H_2) = \frac{10 \ g}{2 \ g/mol} = 5 \ mol$.
$n(O_2) = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$.
Identify the Limiting Reagent $(L.R.)$:
According to the stoichiometry,$2 \ mol$ of $H_2$ requires $1 \ mol$ of $O_2$.
So,$5 \ mol$ of $H_2$ would require $2.5 \ mol$ of $O_2$.
Since we only have $2 \ mol$ of $O_2$,$O_2$ is the $L.R.$
Calculate the product:
Since $1 \ mol$ of $O_2$ produces $2 \ mol$ of $H_2O$,$2 \ mol$ of $O_2$ will produce $2 \times 2 = 4 \ mol$ of $H_2O$.