Chemical stoichiometry Questions in English

(C) The reaction of $CO_{2}$ with $Ca(OH)_{2}$ in excess leads to the formation of $CaCO_{3}$ and water: $Ca(OH)_{2} + CO_{2} \rightarrow CaCO_{3} + H_{2}O$.
Millimoles of $Ca(OH)_{2} = 50 \ mL \times 0.5 \ M = 25 \ mmol$.
Since the stoichiometry is $1:1$,millimoles of $CaCO_{3}$ formed $= 25 \ mmol$.
The reaction of $CaCO_{3}$ with $HCl$ is: $CaCO_{3} + 2HCl \rightarrow CaCl_{2} + H_{2}O + CO_{2}$.
The milliequivalents of $CaCO_{3}$ $= \text{millimoles} \times n\text{-factor} = 25 \times 2 = 50 \ meq$.
At neutralization,milliequivalents of $HCl = \text{milliequivalents of } CaCO_{3} = 50 \ meq$.
Volume of $0.1 \ N \ HCl = \frac{50 \ meq}{0.1 \ N} = 500 \ cm^{3}$.

(C) Assume the total volume of the gas mixture is $100 \ mL$.
Since the volume percentage is equal to the mole percentage for ideal gases,we have $25 \ mol$ of He and $75 \ mol$ of $CH_4$.
Molar mass of He $= 4 \ g/mol$.
Molar mass of $CH_4 = 16 \ g/mol$.
Mass of He $= 25 \times 4 = 100 \ g$.
Mass of $CH_4 = 75 \times 16 = 1200 \ g$.
Total mass of the mixture $= 100 + 1200 = 1300 \ g$.
Mass percentage of $CH_4 = \frac{\text{Mass of } CH_4}{\text{Total mass}} \times 100 = \frac{1200}{1300} \times 100 \approx 92.3 \%$.
Thus,the percentage by mass of methane is approximately $92 \%$.

(B) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $y$.
Given the total number of $Fe$ ions is $x + y = 0.93$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
$2x + 3y = 2$ (charge on $O^{2-}$ is $-2$).
From the first equation,$x = 0.93 - y$.
Substituting into the second equation: $2(0.93 - y) + 3y = 2$.
$1.86 - 2y + 3y = 2$.
$y = 2 - 1.86 = 0.14$.
So,$x = 0.93 - 0.14 = 0.79$.
The percentage of $Fe^{3+}$ ions is $\frac{y}{x+y} \times 100 = \frac{0.14}{0.93} \times 100 \approx 15.05\%$.
Thus,the percentage is nearly $15\%$.

(A) Molar mass of benzene $= 78 \ g/mol$.
Mole of benzene $= \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$.
The reaction is $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_8H_8O + HCl$.
According to the stoichiometry,$1 \ mol$ of benzene produces $1 \ mol$ of $C_8H_8O$.
Therefore,$0.1 \ mol$ of benzene should theoretically produce $0.1 \ mol$ of $C_8H_8O$.
Molar mass of $C_8H_8O = (8 \times 12) + (8 \times 1) + (1 \times 16) = 96 + 8 + 16 = 120 \ g/mol$.
Theoretical yield $= 0.1 \ mol \times 120 \ g/mol = 12 \ g$.
Percentage yield $= \frac{\text{Experimental yield}}{\text{Theoretical yield}} \times 100 = \frac{8.4 \ g}{12 \ g} \times 100 = 70\%$.

(C) The molecular formula of benzene is $C_6H_6$.
The balanced chemical equation for the complete combustion of benzene is:
$C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l)$
From the stoichiometry of the reaction,$1 \text{ mole}$ of benzene produces $6 \text{ moles}$ of carbon dioxide $(CO_2)$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Therefore,the mass of $6 \text{ moles}$ of $CO_2$ is $6 \times 44 \ g = 264 \ g$.
Thus,the complete combustion of one mole of benzene produces $264 \ g$ of carbon dioxide.

(A) The balanced chemical equation for the reaction of fluorine with hot concentrated $KOH$ is:
$2 F_2 + 4 KOH \longrightarrow 4 KF + 2 H_2O + O_2$
From the stoichiometry of the reaction,the molar ratio of $KF : H_2O : O_2$ is $4 : 2 : 1$.
Dividing by $2$,we get the ratio $2 : 1 : 0.5$.

(A) The balanced chemical equation for the reaction between $H_2O_2$ and acidified $KMnO_4$ is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \longrightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
From the stoichiometry,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } H_2O_2$.
Moles of $KMnO_4 = \text{Molarity} \times \text{Volume (in L)} = 0.02 \times 0.5 = 0.01 \text{ mol}$.
Moles of $H_2O_2$ required $= \frac{5}{2} \times 0.01 = 0.025 \text{ mol}$.
Normality of $20 \text{ vol } H_2O_2 = \frac{20}{5.6} \approx 3.57 \text{ N}$.
Since the $n$-factor for $H_2O_2$ in this reaction is $2$,Molarity of $H_2O_2 = \frac{3.57}{2} = 1.785 \text{ M}$.
Volume of $H_2O_2 = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.025}{1.785} \approx 0.014 \text{ L} = 14 \text{ mL}$.
Alternatively,using the law of equivalence: $N_1V_1 = N_2V_2$.
Normality of $KMnO_4 = 0.02 \times 5 = 0.1 \text{ N}$.
$3.57 \times V_1 = 0.1 \times 500 \implies V_1 = \frac{50}{3.57} \approx 14 \text{ mL}$.
Thus,option $(A)$ is the correct answer.

(C) Given:
Volume of $HCl$ solution $(V_1) = 30 \ mL$
Volume of sodium carbonate solution $(Na_2CO_3)$ $(V_2) = 20 \ mL$
Concentration of $Na_2CO_3$ solution $(M_2) = 0.1 \ M$
Volume of $NaOH$ solution $(V_3) = 30.0 \ mL$
Concentration of $NaOH$ solution $(M_3) = 0.2 \ M$
For the reaction between $HCl$ and $Na_2CO_3$ $(Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2)$:
$n_{HCl} = 2 \times n_{Na_2CO_3}$
$M_1V_1 = 2 \times M_2V_2$
$M_1 = \frac{2 \times M_2 \times V_2}{V_1} = \frac{2 \times 0.1 \times 20}{30} = \frac{4}{30} \ M$
Also,for the reaction between $HCl$ and $NaOH$ $(NaOH + HCl \rightarrow NaCl + H_2O)$:
$M_1 V_f = M_3 V_3$ (where $V_f$ is the volume of $HCl$ solution required)
$V_f = \frac{M_3 \times V_3}{M_1} = \frac{0.2 \times 30}{4 / 30} = \frac{6}{4 / 30} = \frac{180}{4} = 90 \ mL$
Hence,option $C$ is the correct answer.

(C) The reaction of metallic sodium with water is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$.
Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$.
Since $2 \ mol$ of $Na$ produce $2 \ mol$ of $NaOH$,$0.02 \ mol$ of $Na$ will produce $0.02 \ mol$ of $NaOH$.
The neutralization reaction is: $NaOH + HCl \rightarrow NaCl + H_2O$.
To neutralize $0.02 \ mol$ of $NaOH$,we require $0.02 \ mol$ of $HCl$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Concentration of $HCl$ solution = $73 \ g/L = \frac{73 \ g/L}{36.5 \ g/mol} = 2 \ M$.
Volume of $HCl$ required = $\frac{\text{moles}}{\text{molarity}} = \frac{0.02 \ mol}{2 \ mol/L} = 0.01 \ L = 10 \ mL$.

(D) Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.96 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
The charge of $O^{2-}$ is $-2$.
Therefore,$2x + 3(0.96 - x) - 2 = 0$.
$2x + 2.88 - 3x - 2 = 0$.
$-x + 0.88 = 0$.
$x = 0.88$.
So,the number of $Fe^{2+}$ ions is $0.88$ and the number of $Fe^{3+}$ ions is $0.96 - 0.88 = 0.08$.
The mole fraction of $Fe^{2+}$ is the ratio of the number of $Fe^{2+}$ ions to the total number of $Fe$ ions.
Mole fraction of $Fe^{2+} = \frac{0.88}{0.96} = \frac{88}{96} = \frac{11}{12}$.

(A) The thermal decomposition of potassium permanganate is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow$.
Molar mass of $KMnO_4 = 39 + 55 + (16 \times 4) = 158 \text{ g/mol}$.
Mass of $0.1 \text{ mole}$ of $KMnO_4 = 0.1 \times 158 = 15.8 \text{ g}$.
From the reaction stoichiometry,$2 \text{ moles}$ of $KMnO_4$ produce $1 \text{ mole}$ of $O_2$ gas.
So,$0.1 \text{ mole}$ of $KMnO_4$ produces $0.05 \text{ mole}$ of $O_2$.
Mass of $O_2$ gas $= 0.05 \times 32 = 1.6 \text{ g}$.
Weight of residue = Initial weight - Weight of $O_2$ gas $= 15.8 - 1.6 = 14.2 \text{ g}$.

(C) Let the mass of $CaCO_3$ be $x \ g$ and the mass of $MgCO_3$ be $(1.84 - x) \ g$.
Heating the carbonates results in the following reactions:
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
$MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$
The molar mass of $CaCO_3 = 100 \ g/mol$,$CaO = 56 \ g/mol$,$MgCO_3 = 84 \ g/mol$,and $MgO = 40 \ g/mol$.
The mass of the residue $(CaO + MgO)$ is $0.96 \ g$.
Mass of $CaO = (x / 100) \times 56 = 0.56x$.
Mass of $MgO = ((1.84 - x) / 84) \times 40 = 0.476(1.84 - x)$.
$0.56x + 0.476(1.84 - x) = 0.96$.
$0.56x + 0.876 - 0.476x = 0.96$.
$0.084x = 0.084$.
$x = 1 \ g$.
Percentage of $CaCO_3 = (1 / 1.84) \times 100 \approx 54.34\%$.

(D) $1$. Calculate total moles of the mixture $(n_{total})$ using the ideal gas law: $PV = nRT$. Given $P = 1 \ atm$,$V = 28 \ L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 273 \ K$. $n_{total} = \frac{1 \times 28}{0.0821 \times 273} \approx 1.25 \ mol$.
$2$. Let $x$ be the moles of $C_2H_6$ and $y$ be the moles of $C_2H_4$. So,$x + y = 1.25$.
$3$. Combustion reactions:
$C_2H_6 + 3.5O_2 \rightarrow 2CO_2 + 3H_2O$
$C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$
$4$. Moles of $O_2$ required: $3.5x + 3y = \frac{128}{32} = 4$.
$5$. Solve the system of equations: $x + y = 1.25 \Rightarrow x = 1.25 - y$.
$6$. Substitute into the second equation: $3.5(1.25 - y) + 3y = 4$ $\Rightarrow 4.375 - 3.5y + 3y = 4$ $\Rightarrow 0.5y = 0.375$ $\Rightarrow y = 0.75$.
$7$. Mole fraction of $C_2H_4 = \frac{y}{x+y} = \frac{0.75}{1.25} = 0.6$.

(C) Let the mass of $NaCl$ be $x \ g$ and the mass of $NaBr$ be $(4.0 - x) \ g$.
The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.
The molar mass of $NaBr = 23 + 80 = 103 \ g/mol$.
The mass of $Na$ in $NaCl = x \times (23 / 58.5) \ g$.
The mass of $Na$ in $NaBr = (4.0 - x) \times (23 / 103) \ g$.
Total mass of $Na = 30 \% \text{ of } 4.0 \ g = 1.2 \ g$.
So,$x(23 / 58.5) + (4.0 - x)(23 / 103) = 1.2$.
Dividing by $23$: $x / 58.5 + (4.0 - x) / 103 = 1.2 / 23 \approx 0.05217$.
$0.01709x + 0.03883 - 0.00971x = 0.05217$.
$0.00738x = 0.01334$.
$x \approx 1.807 \ g$.
Percentage of $NaCl = (1.807 / 4.0) \times 100 \approx 45.17 \%$.
The closest option is $45 \%$.

(A) $1$. Determine the empirical formula:
$C = 92.3 \%, H = 7.7 \%$.
Moles of $C = \frac{92.3}{12} = 7.69$,Moles of $H = \frac{7.7}{1} = 7.7$.
Ratio $C:H = 1:1$. Empirical formula is $CH$.
$2$. Calculate moles of hydrocarbon:
Given mass $= 39 \ g$. Molar mass of $CH = 13 \ g/mol$.
Moles of $CH = \frac{39}{13} = 3 \ mol$.
$3$. Determine moles of water $(x)$:
Reaction: $2H_2O + 2Na \rightarrow 2NaOH + H_2$.
$2 \ mol$ of $H_2O$ produces $1 \ mol$ of $H_2$.
To produce $0.75 \ mol$ of $H_2$,we need $x = 0.75 \times 2 = 1.5 \ mol$ of $H_2O$.
$4$. Combustion reaction:
For $CH$ (or $C_2H_2$ unit),the combustion is $C_2H_2 + 2.5 O_2 \rightarrow 2 CO_2 + H_2O$.
Since we have $3 \ mol$ of $CH$ (which is $1.5 \ mol$ of $C_2H_2$),the reaction is:
$1.5 C_2H_2 + 3.75 O_2 \rightarrow 3 CO_2 + 1.5 H_2O$.
$5$. Calculate mass of $O_2$:
Moles of $O_2 = 3.75 \ mol$.
Mass of $O_2 = 3.75 \times 32 = 120 \ g$.

(B) The reaction between $CO_2$ and $MOH$ is: $CO_2 + MOH \rightarrow MHCO_3$.
Since $0.2 \ mol$ of $MOH$ reacts completely,$0.2 \ mol$ of $CO_2$ is produced.
The thermal decomposition of metal hydrogen carbonate is: $2MHCO_3 \xrightarrow{\Delta} M_2CO_3 + CO_2 + H_2O$.
From the stoichiometry,$1 \ mol$ of $CO_2$ is produced from $2 \ mol$ of $MHCO_3$.
Therefore,$0.2 \ mol$ of $CO_2$ is produced from $2 \times 0.2 = 0.4 \ mol$ of $MHCO_3$.
The mass '$x$' is calculated as: $x = \text{moles} \times \text{molar mass} = 0.4 \ mol \times 84 \ g \ mol^{-1} = 33.6 \ g$.

(D) $1$. Calculate the mass of pure $H_2SO_4$ required:
$n = M \times V(L) = 0.5 \ mol \ L^{-1} \times 0.5 \ L = 0.25 \ mol$.
Mass of $H_2SO_4 = n \times \text{molar mass} = 0.25 \ mol \times 98.079 \ g \ mol^{-1} = 24.51975 \ g \approx 24.52 \ g$.
$2$. Calculate the mass of the $90\%$ solution required:
Since the solution is $90\%$ pure,$\text{Mass of solution} = \frac{\text{Mass of solute}}{\text{Percentage}} = \frac{24.51975 \ g}{0.90} = 27.244 \ g \approx 27.24 \ g$.

(C) The balanced chemical equation for the formation of ammonia is:
$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$
According to the stoichiometry:
$1 \ mol$ of $N_2$ ($22.4 \ L$ at $NTP$) reacts with $3 \ mol$ of $H_2$ ($3 \times 22.4 \ L = 67.2 \ L$ at $NTP$) to produce $2 \ mol$ of $NH_3$ ($2 \times 22.4 \ L = 44.8 \ L$ at $NTP$).
The molar mass of $NH_3$ is $17 \ g/mol$. Therefore,$2 \ mol$ of $NH_3$ corresponds to $2 \times 17 = 34 \ g$.
For $34 \ g$ of $NH_3$,we require $67.2 \ L$ of $H_2$ and $22.4 \ L$ of $N_2$.
For $3.40 \ g$ of $NH_3$ (which is $0.1$ times $34 \ g$),the required volumes are:
Volume of $H_2 = 0.1 \times 67.2 \ L = 6.72 \ L$
Volume of $N_2 = 0.1 \times 22.4 \ L = 2.24 \ L$
Thus,the required volumes are $6.72 \ L$ of $H_2$ and $2.24 \ L$ of $N_2$.

(B) The balanced chemical equation for the formation of ammonia is:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
From the stoichiometry of the reaction:
$2 \text{ moles}$ of $NH_3$ are produced from $3 \text{ moles}$ of $H_2$.
Molar mass of $NH_3 = 17 \text{ g/mol}$.
Molar mass of $H_2 = 2 \text{ g/mol}$.
Therefore,$2 \times 17 \text{ g} = 34 \text{ g}$ of $NH_3$ is produced from $3 \times 2 \text{ g} = 6 \text{ g}$ of $H_2$.
To produce $100 \text{ g}$ of $NH_3$,the mass of $H_2$ required is:
$\text{Mass of } H_2 = \frac{6 \text{ g}}{34 \text{ g}} \times 100 \text{ g} = 17.647 \text{ g} \approx 17.65 \text{ g}$.

(C) The balanced chemical equation is: $Al(OH)_3 + 3HCl \longrightarrow AlCl_3 + 3H_2O$.
The molar mass of $Al(OH)_3 = 27 + 3 \times (16 + 1) = 78 \ g/mol$.
The molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Total mass of $HCl$ produced per day $= 3.0 \ g/L \times 2.6 \ L = 7.8 \ g$.
From the stoichiometry,$3 \times 36.5 = 109.5 \ g$ of $HCl$ is neutralized by $78 \ g$ of $Al(OH)_3$.
Therefore,$7.8 \ g$ of $HCl$ is neutralized by $\frac{78}{109.5} \times 7.8 = 5.556 \ g$ of $Al(OH)_3$.
Each tablet contains $400 \ mg = 0.4 \ g$ of $Al(OH)_3$.
Number of tablets required $= \frac{5.556 \ g}{0.4 \ g/tablet} = 13.89 \approx 14$ tablets.

(C) The balanced chemical equation for the combustion of methane is:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Molar mass of $CH_4 = 12 + 4(1) = 16 \text{ g/mol}$
Molar mass of $CO_2 = 12 + 2(16) = 44 \text{ g/mol}$
From the stoichiometry of the reaction,$16 \text{ g}$ of $CH_4$ produces $44 \text{ g}$ of $CO_2$.
Therefore,$32 \text{ g}$ of $CH_4$ will produce:
$\text{Amount of } CO_2 = \frac{44 \times 32}{16} = 88 \text{ g}$

(A) The balanced chemical equation is: $3 NO_2 + H_2O \longrightarrow 2 HNO_3 + NO$.
From the stoichiometry of the reaction,$3$ moles of $NO_2$ produce $2$ moles of $HNO_3$.
To produce $4$ moles of $HNO_3$,we need:
$\text{Moles of } NO_2 = \frac{3}{2} \times 4 = 6 \text{ moles}$.
The molar mass of $NO_2 = 14 + (2 \times 16) = 46 \ g/mol$.
Therefore,the mass of $NO_2$ required $= 6 \text{ moles} \times 46 \ g/mol = 276 \ g$.
Hence,the correct option is $A$.

(C) Initial volume $(V_1) = 50 \ mL$
Initial normality $(N_1) = 0.1 \ N$
Final volume $(V_2) = 50 \ mL + 150 \ mL = 200 \ mL$
Using the dilution formula: $N_1 \times V_1 = N_2 \times V_2$
$0.1 \times 50 = N_2 \times 200$
$N_2 = \frac{0.1 \times 50}{200} = 0.025 \ N = \frac{1}{40} \ N$
For $Na_2CO_3$,the valence factor $(Z)$ is $2$ (total positive charge of $Na^+$ ions).
Relation between Normality and Molarity: $N = M \times Z$
$M = \frac{N}{Z} = \frac{1/40}{2} = \frac{1}{80} \ M$
Therefore,the molarity of the resultant solution is $M/80$.

(C) Given,
Molarity of $HCl$ $(M_1) = 0.1 \ M$
Volume of $Na_2CO_3$ solution $(V_2) = 20 \ mL$
Mass of $Na_2CO_3 = 0.106 \ g$
Molar mass of $Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$
Moles of $Na_2CO_3 = \frac{0.106 \ g}{106 \ g/mol} = 0.001 \ mol$
Molarity of $Na_2CO_3$ $(M_2) = \frac{0.001 \ mol}{0.020 \ L} = 0.05 \ M$
The neutralization reaction is: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$
Using the equivalence relation: $n_{HCl} = 2 \times n_{Na_2CO_3}$
$M_1 \times V_1 = 2 \times (M_2 \times V_2)$
$0.1 \times V_1 = 2 \times (0.05 \times 20)$
$0.1 \times V_1 = 2 \times 1 = 2$
$V_1 = \frac{2}{0.1} = 20 \ mL$

(A) $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. $100 \ g \ CaCO_3$ gives $22.4 \ L \ CO_2$ at $STP$. So,$10 \ g \ CaCO_3$ gives $\frac{22.4 \times 10}{100} = 2.24 \ L \ CO_2$ $(iv)$.
$(B)$ $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. $106 \ g \ Na_2CO_3$ gives $22.4 \ L \ CO_2$. So,$1.06 \ g \ Na_2CO_3$ gives $\frac{22.4 \times 1.06}{106} = 0.224 \ L \ CO_2$ $(i)$.
$(C)$ $C + O_2 \rightarrow CO_2$. $12 \ g \ C$ gives $22.4 \ L \ CO_2$. So,$2.4 \ g \ C$ gives $\frac{22.4 \times 2.4}{12} = 4.48 \ L \ CO_2$ $(ii)$.
$(D)$ $2CO + O_2 \rightarrow 2CO_2$. $56 \ g \ CO$ gives $2 \times 22.4 \ L \ CO_2 = 44.8 \ L \ CO_2$. So,$0.56 \ g \ CO$ gives $\frac{44.8 \times 0.56}{56} = 0.448 \ L \ CO_2$ $(iii)$.
Thus,the correct match is $A-iv, B-i, C-ii, D-iii$.

(B) The balanced chemical equation for the oxidation of carbon monoxide is:
$CO(g) + \frac{1}{2} O_2(g) \longrightarrow CO_2(g)$
According to the stoichiometry of the reaction,$1 \text{ mole}$ of $CO$ produces $1 \text{ mole}$ of $CO_2$.
At $STP$,$1 \text{ mole}$ of any ideal gas occupies $22.414 \ L$.
Therefore,$22.414 \ L$ of $CO$ produces $22.414 \ L$ of $CO_2$.
Since the volume of $CO_2$ formed is $11.207 \ L$,the volume of $CO$ $(X)$ required is also $11.207 \ L$ because the molar ratio is $1:1$.

(B) The combustion reaction of methane is: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$.
From the stoichiometry,$1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$.
The reaction of $CO_2$ with calcium hydroxide is: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O$.
The molar mass of $Ca(OH)_2 = 40 + 2 \times (16 + 1) = 74 \ g/mol$.
The molar mass of $CaCO_3 = 40 + 12 + 3 \times 16 = 100 \ g/mol$.
Given $500 \ g$ of $CaCO_3$ precipitate is formed,which corresponds to $500 / 100 = 5 \ moles$ of $CaCO_3$.
Since $1 \ mole$ of $CaCO_3$ is produced from $1 \ mole$ of $CO_2$,$5 \ moles$ of $CO_2$ must have been produced.
From the combustion reaction,$5 \ moles$ of $CO_2$ are produced from $5 \ moles$ of $CH_4$.
The molar mass of $CH_4 = 12 + 4 \times 1 = 16 \ g/mol$.
Therefore,the mass of $CH_4$ $(x)$ = $5 \ moles \times 16 \ g/mol = 80 \ g$.

(B) We know that concentration is dependent on the number of moles of solute and the volume of solution in $L$ and it is given by:
Molarity $= \frac{\text{number of moles of solute}}{\text{volume of solution in } L}$
Here,$n = 0.2 \ mol$,$V = 250 \ mL = 0.25 \ L$
Therefore,
Molarity $= \frac{0.2}{0.25} = 0.8 \ M$
Hence,the concentration of the solution containing $0.2 \ mol$ of sulphuric acid in $250 \ mL$ of water will be $0.8 \ M$.

(A) The dissociation of $CaCl_2$ is given by: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$.
One mole of $CaCl_2$ produces $2$ moles of chloride ions $(Cl^-)$.
Number of moles of $Cl^-$ ions $= \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}} = 0.05 \ mol$.
Since $2$ moles of $Cl^-$ come from $1$ mole of $CaCl_2$,the moles of $CaCl_2 = \frac{0.05}{2} = 0.025 \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.025 \ mol}{0.5 \ L} = 0.05 \ M$.
Thus,the molarity of the solution is $0.05 \ M$.

(C) Molar mass of oxalic acid $(COOH)_2 \cdot 2H_2O = 126 \ g/mol$.
Equivalent mass of oxalic acid = $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{126}{2} = 63 \ g/eq$.
Number of gram equivalents = $\frac{\text{Given mass}}{\text{Equivalent mass}} = \frac{0.63}{63} = 0.01 \ eq$.
Normality $(N) = \frac{\text{Number of gram equivalents}}{\text{Volume of solution in } L} = \frac{0.01}{250/1000} = \frac{0.01 \times 1000}{250} = 0.04 \ N$.
Therefore,the normality of the solution is $0.04 \ N$.

(C) Given,density $(d)$ of $HCl$ solution $= 1.2 \ g \ mL^{-1}$.
Percentage strength by mass $(w/w)$ $= 40 \%$.
Molar mass of $HCl$ $(M_{HCl})$ $= 1 + 35.5 = 36.5 \ g \ mol^{-1}$.
The formula for molarity is: $\text{Molarity} = \frac{\text{Percentage}(w/w) \times d \times 10}{M_{HCl}}$.
Substituting the values: $\text{Molarity} = \frac{40 \times 1.2 \times 10}{36.5} = \frac{480}{36.5} \approx 13.15 \ M$.
Therefore,the molarity is nearly $13 \ M$.

(C) The $n$-factor for $Na_2CO_3$ is $2$ because it acts as a base and can accept $2 \ H^+$ ions.
Normality is related to molarity by the formula: $\text{Normality} = n\text{-factor} \times \text{Molarity}$.
Given: $\text{Normality} = 0.2 \ N$ and $n\text{-factor} = 2$.
Therefore,$\text{Molarity} = \frac{\text{Normality}}{n\text{-factor}} = \frac{0.2}{2} = 0.1 \ M$.
Hence,the correct option is $C$.

(B) First,calculate the molarity $(M_1)$ of the initial $250 \ mL$ solution:
Molar mass of $Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$.
$M_1 = \frac{\text{mass}}{\text{molar mass} \times \text{volume in L}} = \frac{2.65 \ g}{106 \ g/mol \times 0.250 \ L} = 0.1 \ M$.
Now,use the dilution formula $M_1V_1 = M_2V_2$:
$0.1 \ M \times 10 \ mL = M_2 \times 1000 \ mL$.
$M_2 = \frac{0.1 \times 10}{1000} = 0.001 \ M$.

(A) The number of moles of $N_2$ is $n_{N_2} = \frac{5 \ g}{28 \ g/mol} \approx 0.1786 \ mol$.
The number of moles of $Ar$ is $n_{Ar} = \frac{6 \ g}{40 \ g/mol} = 0.15 \ mol$.
The mole fraction of $N_2$ is $\chi_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{Ar}} = \frac{0.1786}{0.1786 + 0.15} = \frac{0.1786}{0.3286} \approx 0.5435$.
The partial pressure of $N_2$ is $p_{N_2} = \chi_{N_2} \times p_{Total} = 0.5435 \times 30 \ bar \approx 16.305 \ bar$.
Rounding to the nearest provided option,the value is $16.36 \ bar$.

(B) The chemical reaction is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
Initially,we have $1 \ mole$ of $N_2$ and $3 \ moles$ of $H_2$.
Let $x$ be the moles of $N_2$ reacted. According to stoichiometry,$3x$ moles of $H_2$ react to form $2x$ moles of $NH_3$.
Given that $2x = 1 \ mole$ of $NH_3$ is formed,so $x = 0.5 \ mole$.
At equilibrium,the moles of each gas are:
$n(N_2) = 1 - 0.5 = 0.5 \ mole$
$n(H_2) = 3 - 3(0.5) = 1.5 \ moles$
$n(NH_3) = 1 \ mole$
Total moles $n_{total} = 0.5 + 1.5 + 1 = 3 \ moles$.
Using Dalton's Law,the partial pressure of $N_2$ is $P(N_2) = (n(N_2) / n_{total}) \times P_{total}$.
$P(N_2) = (0.5 / 3) \times 15 \ atm = 0.5 \times 5 = 2.5 \ atm$.

(A) The chemical equation for the decomposition of sodium azide is: $2 NaN_{3(s)} \rightarrow 2 Na_{(s)} + 3 N_{2(g)}$.
Given mass of $NaN_3 = 130 \ g$. Molar mass of $NaN_3 = 23 + (3 \times 14) = 65 \ g \ mol^{-1}$.
Number of moles of $NaN_3 = \frac{130 \ g}{65 \ g \ mol^{-1}} = 2 \ mol$.
According to the stoichiometry,$2 \ mol$ of $NaN_3$ produces $3 \ mol$ of $N_2$ gas.
Using the ideal gas equation $PV = nRT$,where $n = 3 \ mol$,$V = 10 \ L$,$T = 300 \ K$,and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$:
$P = \frac{nRT}{V} = \frac{3 \times 0.082 \times 300}{10} = \frac{73.8}{10} = 7.38 \ atm$.

(A) Synthesis gas is a mixture of $CO$ and $H_2$.
To produce methanol $(CH_3OH)$,the chemical reaction is: $CO(g) + 2H_2(g) \longrightarrow CH_3OH(g)$.
According to the stoichiometry of the reaction,$1$ mole of $CO$ reacts with $2$ moles of $H_2$.
Therefore,the ratio of $CO:H_2$ in the synthesis gas should be $1: 2$.
Hence,option $(A)$ is correct.

(A) $200 \ ppm$ means $200 \ g$ of $CaCO_3$ in $10^6 \ g$ (or approximately $10^6 \ mL$) of water.
$\text{Molar mass of } CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
$\text{Molarity} = \frac{\text{mass in } g}{\text{Molar mass}} \times \frac{1000}{V_{mL}} = \frac{200}{100} \times \frac{1000}{10^6} = 2 \times 10^{-3} \ M$.
$\text{Equivalent mass of } CaCO_3 = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{100}{2} = 50 \ g/eq$.
$\text{Normality} = \frac{\text{mass in } g}{\text{Equivalent mass}} \times \frac{1000}{V_{mL}} = \frac{200}{50} \times \frac{1000}{10^6} = 4 \times 10^{-3} \ N$.

(C) The balanced chemical equation for the reaction of ammonia with excess chlorine is:
$NH_3 + 3Cl_2 \longrightarrow NCl_3 + 3HCl$
From the stoichiometry of the balanced equation,$1 \text{ mole of } NH_3$ reacts with $3 \text{ moles of } Cl_2$.
Therefore,the mole ratio of $NH_3$ to $Cl_2$ is $1: 3$.

(B) The balanced chemical equation for the reaction of fluorine with hot concentrated $KOH$ is:
$2F_2 + 4KOH \longrightarrow 4KF + 2H_2O + O_2$
Dividing the entire equation by $2$ to match the stoichiometry for $1 \text{ mole}$ of $F_2$:
$F_2 + 2KOH \longrightarrow 2KF + H_2O + 0.5O_2$
From the balanced equation,the molar ratio of $KF : H_2O : O_2$ is $2 : 1 : 0.5$.

(B) The normality $(N)$ of a solution is related to its molarity $(M)$ by the formula: $N = M \times \text{n-factor}$.
For $H_2SO_4$, the n-factor (basicity) is $2$.
Given $N = 1 \,N$, we have $1 = M \times 2$, which implies $M = 0.5 \,M$.
We know that Molarity $(M)$ is defined as the number of moles $(n)$ per volume $(V)$ in liters: $M = \frac{n}{V(L)}$.
Substituting the given values: $0.5 = \frac{0.2}{V(L)}$.
Solving for $V(L)$: $V(L) = \frac{0.2}{0.5} = 0.4 \,L$.
Converting to milliliters: $0.4 \,L \times 1000 \,mL/L = 400 \,mL$.

(B) The balanced chemical equation for the reaction is:
$Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Moles of ferrous ammonium sulphate $= M \times V = 0.1 \ M \times 2 \ L = 0.2 \ mol$.
From the stoichiometry,$1 \ mole$ of $K_2Cr_2O_7$ reacts with $6 \ moles$ of $Fe^{2+}$.
Moles of $K_2Cr_2O_7$ required $= \frac{0.2}{6} = \frac{0.1}{3} \ mol$.
Molarity $x = \frac{n}{V} = \frac{0.1/3 \ mol}{1/3 \ L} = 0.1 \ M$.
Thus,option $(B)$ is correct.

(B) Phosphorous acid $(H_3PO_3)$ is a dibasic acid,meaning it provides $2$ moles of $H^+$ ions per mole of acid.
The normality of the $H_3PO_3$ solution is calculated as: $\text{Normality} = \text{Molarity} \times \text{Basicity} = 0.3 \ M \times 2 = 0.6 \ N$.
Using the neutralization principle $N_1 V_1 = N_2 V_2$:
For $NaOH$,$N_1 = 0.1 \ N$ and for $H_3PO_3$,$N_2 = 0.6 \ N$.
$0.1 \times V_1 = 0.6 \times 100$.
$V_1 = \frac{0.6 \times 100}{0.1} = 600 \ mL$.

(B) The balanced chemical equation for the combustion of glucose is:
$C_6H_{12}O_6 + 6O_2 \longrightarrow 6CO_2 + 6H_2O$
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.
From the stoichiometry,$1 \ mol$ of glucose $(180 \ g)$ produces $6 \ mol$ of $CO_2$ $(6 \times 44 = 264 \ g)$ and $6 \ mol$ of $H_2O$ $(6 \times 18 = 108 \ g)$.
For $10 \ g$ of glucose:
Mass of $CO_2 = \frac{264 \ g}{180 \ g} \times 10 \ g = 14.66 \ g$.
Mass of $H_2O = \frac{108 \ g}{180 \ g} \times 10 \ g = 6.0 \ g$.