Chemical equation and limiting reagent Questions in English

(B) The given chemical equation is: $aCaCO_3 + bH_3PO_4 \rightarrow pCa_3(PO_4)_2 + qCO_2 + rH_2O$
To balance the equation,first balance the $Ca$ and $PO_4$ groups: $3CaCO_3 + 2H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 + qCO_2 + rH_2O$
Next,balance the carbon atoms: $3CaCO_3 + 2H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 + 3CO_2 + rH_2O$
Finally,balance the hydrogen and oxygen atoms: $3CaCO_3 + 2H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 + 3CO_2 + 3H_2O$
Comparing this with the given equation,we get $a = 3$ and $b = 2$.
Therefore,the ratio $a / b = 3 / 2$.

(B) To balance the equation $x KNO_3 + y C_{12}H_{22}O_{11} \longrightarrow p N_2 + q CO_2 + r H_2O + s K_2CO_3$:
$1$. Balance $K$: $x = 2s$
$2$. Balance $N$: $x = 2p$
$3$. Balance $C$: $12y = q + s$
$4$. Balance $H$: $22y = 2r \implies 11y = r$
$5$. Balance $O$: $3x + 11y = 2q + r + 3s$
Substituting the values from option $(b)$: $x=48, y=5, p=24, q=36, r=55, s=24$.
$48 KNO_3 + 5 C_{12}H_{22}O_{11} \longrightarrow 24 N_2 + 36 CO_2 + 55 H_2O + 24 K_2CO_3$
Checking atoms:
$K: 48 = 2(24) = 48$ (Balanced)
$N: 48 = 2(24) = 48$ (Balanced)
$C: 12(5) = 60; 36 + 24 = 60$ (Balanced)
$H: 22(5) = 110; 55(2) = 110$ (Balanced)
$O: 3(48) + 11(5) = 144 + 55 = 199; 2(36) + 55 + 3(24) = 72 + 55 + 72 = 199$ (Balanced)
Thus,option $(b)$ is correct.

(D) The balanced chemical equation is:
$CaO + 2HNO_3 \longrightarrow Ca(NO_3)_2 + H_2O$
Molar masses:
$CaO = 40 + 16 = 56 \ g/mol$
$HNO_3 = 1 + 14 + 48 = 63 \ g/mol$
$Ca(NO_3)_2 = 40 + 2(14 + 48) = 40 + 124 = 164 \ g/mol$
Given amounts:
$CaO = 56 \ g = 1 \ mol$
$HNO_3 = 63 \ g = 1 \ mol$
According to the stoichiometry,$1 \ mol$ of $CaO$ requires $2 \ mol$ of $HNO_3$. Since we only have $1 \ mol$ of $HNO_3$,$HNO_3$ is the limiting reagent.
From the equation,$2 \ mol$ of $HNO_3$ produces $1 \ mol$ of $Ca(NO_3)_2$ $(164 \ g)$.
Therefore,$1 \ mol$ of $HNO_3$ will produce:
$\frac{164 \ g}{2} = 82 \ g$ of $Ca(NO_3)_2$.

(B) The balanced chemical equation is:
$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$
Given mass of $MnO_2 = 100 \ g$.
Moles of $MnO_2 = \frac{100 \ g}{86.9 \ g/mol} = 1.15 \ mol$.
Given mass of $HCl = 100 \ g$.
Moles of $HCl = \frac{100 \ g}{36.5 \ g/mol} = 2.74 \ mol$.
From the stoichiometry,$1 \ mol$ of $MnO_2$ reacts with $4 \ mol$ of $HCl$.
Therefore,$1.15 \ mol$ of $MnO_2$ require $1.15 \times 4 = 4.6 \ mol$ of $HCl$.
Since the available moles of $HCl$ $(2.74 \ mol)$ are less than the required moles $(4.6 \ mol)$,$HCl$ is the limiting reagent.

(C) The balanced chemical equation is: $H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$
Initial moles of $H_2SO_4 = \text{Molarity} \times \text{Volume} = 0.2 \ M \times 1 \ L = 0.2 \ mol$.
Initial moles of $NaOH = \text{Molarity} \times \text{Volume} = 0.2 \ M \times 1 \ L = 0.2 \ mol$.
According to the stoichiometry,$1 \ mol$ of $H_2SO_4$ requires $2 \ mol$ of $NaOH$.
Therefore,$0.2 \ mol$ of $H_2SO_4$ would require $0.4 \ mol$ of $NaOH$.
Since only $0.2 \ mol$ of $NaOH$ is available,$NaOH$ is the limiting reagent.
From the stoichiometry,$2 \ mol$ of $NaOH$ produces $1 \ mol$ of $Na_2SO_4$.
So,$0.2 \ mol$ of $NaOH$ will produce $0.1 \ mol$ of $Na_2SO_4$.
Molar mass of $Na_2SO_4 = (2 \times 23) + 32 + (4 \times 16) = 142 \ g/mol$.
Mass of $Na_2SO_4 = \text{moles} \times \text{molar mass} = 0.1 \ mol \times 142 \ g/mol = 14.2 \ g$.

(B) The combustion reaction for methane is as follows:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Initial moles:
$CH_4 = 10, O_2 = 50, CO_2 = 0, H_2O = 0$
Since $CH_4$ is the limiting reagent,it will be completely consumed.
For $10 \ \text{moles}$ of $CH_4$,$2 \times 10 = 20 \ \text{moles}$ of $O_2$ are consumed.
Final moles:
$O_2 = 50 - 20 = 30 \ \text{moles}$
$CO_2 = 10 \ \text{moles}$
$H_2O = 2 \times 10 = 20 \ \text{moles}$
Thus,the number of moles of $O_2$,$H_2O$,and $CO_2$ are $30, 20, 10$ respectively.
Option $B$ is the correct answer.

(C) First,we balance the chemical equation:
$7 H_{2(g)} + 2 NO_{2(g)} \longrightarrow 2 NH_{3(g)} + 4 H_2O_{(g)}$
From the balanced equation,$2$ moles of $NH_3$ are produced by $7$ moles of $H_2$.
Therefore,$1$ mole of $NH_3$ is produced by $\frac{7}{2}$ moles of $H_2$.
For $10$ moles of $NH_3$,the moles of $H_2$ required are:
$= \frac{7}{2} \times 10 = 35$ moles.
Thus,the correct option is $(C)$.

(A) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
First,calculate the moles of reactants:
$n_{N_2} = \frac{50.0 \ g}{28.0 \ g/mol} \approx 1.786 \ mol$.
$n_{H_2} = \frac{10.0 \ g}{2.016 \ g/mol} \approx 4.96 \ mol$.
According to the stoichiometry,$1 \ mol$ of $N_2$ requires $3 \ mol$ of $H_2$.
For $1.786 \ mol$ of $N_2$,we need $1.786 \times 3 = 5.358 \ mol$ of $H_2$.
Since we only have $4.96 \ mol$ of $H_2$,$H_2$ is the limiting reagent.
The moles of $NH_3$ produced depend on the limiting reagent $(H_2)$:
$n_{NH_3} = n_{H_2} \times \frac{2 \ mol \ NH_3}{3 \ mol \ H_2} = 4.96 \times \frac{2}{3} \approx 3.31 \ mol$.
Using the exact mass calculation: $n_{NH_3} = \frac{10 \ g}{6 \ g} \times 2 = \frac{20}{6} = 3.33 \ mol$.

(B) The reaction between $aniline$ and $acetic$ $anhydride$ is:
$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$
Here,$1$ mole of $aniline$ reacts with $1$ mole of $acetic$ $anhydride$ to form $1$ mole of $acetanilide$ $(C_6H_5NHCOCH_3)$.
Given: $2$ moles of $aniline$ and $1$ mole of $acetic$ $anhydride$.
Since $acetic$ $anhydride$ is the limiting reagent,$1$ mole of $acetanilide$ will be formed.
The molar mass of $acetanilide$ $(C_8H_9NO)$ is $135 \ g/mol$.
Therefore,the yield is $1 \ mol \times 135 \ g/mol = 135 \ g$.

(B) The balanced chemical equation for the reaction is:
$Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)$
Calculate the initial millimoles of each reactant:
Millimoles of $Pb(NO_{3})_{2} = 5 \ mL \times 0.1 \ M = 0.5 \ mmol$
Millimoles of $KI = 10 \ mL \times 0.02 \ M = 0.2 \ mmol$
According to the stoichiometry,$1 \ mol$ of $Pb(NO_{3})_{2}$ reacts with $2 \ mol$ of $KI$. Therefore,$0.2 \ mmol$ of $KI$ will react with $0.1 \ mmol$ of $Pb(NO_{3})_{2}$.
Since $KI$ is the limiting reagent,the amount of $PbI_{2}$ formed depends on the amount of $KI$.
From the stoichiometry,$2 \ mmol$ of $KI$ produces $1 \ mmol$ of $PbI_{2}$.
So,$0.2 \ mmol$ of $KI$ will produce $0.1 \ mmol$ of $PbI_{2}$.
Convert millimoles to moles:
$0.1 \ mmol = 0.1 \times 10^{-3} \ mol = 10^{-4} \ mol$.

(A) The heat of neutralisation for the reaction between a strong acid and a strong base is defined as the heat released when $1 \ mole$ of $H^+$ ions reacts with $1 \ mole$ of $OH^-$ ions to form $1 \ mole$ of $H_2O$,which is $13.7 \ kcal$.
The balanced chemical equation is:
$HCl + NaOH \longrightarrow NaCl + H_2O; \Delta H = -13.7 \ kcal \text{ per mole of } H_2O \text{ formed}$.
Given amounts:
$HCl = 0.6 \ mole$
$NaOH = 0.25 \ mole$
Since $NaOH$ is the limiting reagent,the amount of $H_2O$ formed is determined by the amount of $NaOH$ present,which is $0.25 \ mole$.
Heat released $= \text{Heat of neutralisation} \times \text{moles of } H_2O \text{ formed}$
Heat released $= 13.7 \ kcal/mole \times 0.25 \ mole = 3.425 \ kcal$.

(C) The balanced chemical equation for the reaction of chlorine with cold and dilute $KOH$ is:
$Cl_2 + 2KOH \rightarrow KCl + KClO + H_2O$
Initial moles of $Cl_2 = 1 \text{ mole}$
Initial moles of $KOH = \text{Molarity} \times \text{Volume} = 2 \text{ M} \times 2 \text{ L} = 4 \text{ moles}$
According to the stoichiometry,$1 \text{ mole}$ of $Cl_2$ reacts with $2 \text{ moles}$ of $KOH$.
Moles of $KOH$ reacted $= 2 \text{ moles}$
Moles of $KOH$ remaining $= 4 - 2 = 2 \text{ moles}$
Moles of $Cl^{-}$ formed (from $KCl$) $= 1 \text{ mole}$
Moles of $ClO^{-}$ formed (from $KClO$) $= 1 \text{ mole}$
Final concentrations (Volume $= 2 \text{ L}$):
$[Cl^{-}] = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}$
$[ClO^{-}] = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}$
$[OH^{-}] = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M}$

(A) $1$. Calculate the mass of $HCl$ solution: $\text{Mass} = \text{Density} \times \text{Volume} = 1.13 \ g \ mL^{-1} \times 300 \ mL = 339 \ g$.
$2$. Calculate the mass of pure $HCl$: $\text{Mass of } HCl = 339 \ g \times 0.3855 = 130.68 \ g$.
$3$. Calculate moles of reactants:
Moles of $CaCO_{3} = \frac{90 \ g}{100 \ g \ mol^{-1}} = 0.90 \ mol$.
Moles of $HCl = \frac{130.68 \ g}{36.5 \ g \ mol^{-1}} \approx 3.58 \ mol$.
$4$. Determine the Limiting Reagent $(LR)$:
According to the stoichiometry,$1 \ mol$ of $CaCO_{3}$ reacts with $2 \ mol$ of $HCl$.
For $0.90 \ mol$ of $CaCO_{3}$,we need $0.90 \times 2 = 1.80 \ mol$ of $HCl$.
Since we have $3.58 \ mol$ of $HCl$,$CaCO_{3}$ is the limiting reagent.
$5$. Calculate remaining $HCl$:
$HCl$ consumed $= 1.80 \ mol$.
$HCl$ remaining $= 3.58 - 1.80 = 1.78 \ mol$.
Mass of $HCl$ remaining $= 1.78 \ mol \times 36.5 \ g \ mol^{-1} = 64.97 \ g$.

(C) The chemical equation for the reaction is: $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$.
First,calculate the mass of pure $H_2SO_4$:
Mass of solution $= \text{Volume} \times \text{Density} = 50 \text{ mL} \times 1.3 \text{ g mL}^{-1} = 65 \text{ g}$.
Mass of pure $H_2SO_4 = 65 \text{ g} \times 0.50 = 32.5 \text{ g}$.
Moles of $H_2SO_4 = \frac{32.5 \text{ g}}{98 \text{ g mol}^{-1}} \approx 0.3316 \text{ mol}$.
Moles of $Zn = \frac{20 \text{ g}}{65 \text{ g mol}^{-1}} \approx 0.3077 \text{ mol}$.
Since the stoichiometry of the reaction is $1:1$,$Zn$ is the limiting reagent because it has fewer moles.
Therefore,moles of $H_2$ produced $= \text{moles of } Zn = 0.3077 \text{ mol}$.
Volume of $H_2$ at $STP = 0.3077 \text{ mol} \times 22.4 \text{ L mol}^{-1} \approx 6.892 \text{ L}$.