Chemical stoichiometry Questions in English

(D) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g/mol$.
Number of moles of $CH_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$.
At $STP$,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$ of volume.
Therefore,the volume occupied by $2 \ mol$ of $CH_4 = 2 \times 22.4 \ dm^3 = 44.8 \ dm^3$.

(C) The number of moles of $He$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \ g}{4 \ g \ mol^{-1}} = 0.5 \ mol$.
At $STP$,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$.
Therefore,the volume occupied by $0.5 \ mol$ of $He$ gas is: $V = 0.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 11.2 \ dm^3$.

(B) The molar mass of ethane $(C_2H_6)$ is $(2 \times 12) + (6 \times 1) = 30 \ g/mol$.
Number of moles of ethane $= \frac{\text{mass}}{\text{molar mass}} = \frac{75 \ g}{30 \ g/mol} = 2.5 \ mol$.
At $S$.$T$.$P$.,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$.
Volume occupied $= 2.5 \ mol \times 22.4 \ dm^3/mol = 56 \ dm^3$.

(D) At $S.T.P.$,the molar volume of an ideal gas is $22.4 \ dm^3 \ mol^{-1}$.
Number of moles of methane $(CH_4) = \frac{\text{Given volume}}{\text{Molar volume at } S.T.P.} = \frac{33.6 \ dm^3}{22.4 \ dm^3 \ mol^{-1}} = 1.5 \ mol$.
The molar mass of methane $(CH_4) = 12 + (4 \times 1) = 16 \ g \ mol^{-1}$.
Mass of methane $= \text{moles} \times \text{molar mass} = 1.5 \ mol \times 16 \ g \ mol^{-1} = 24 \ g$.
Converting to $kg$: $24 \ g = 24 \times 10^{-3} \ kg = 2.4 \times 10^{-2} \ kg$.

(D) The balanced chemical equation is: $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$
From the stoichiometry of the reaction,$3 \ moles$ of $O_2$ are produced along with $2 \ moles$ of $KCl$.
At $S.T.P.$,$1 \ mole$ of any gas occupies $22.4 \ L$.
Thus,$3 \ moles$ of $O_2$ occupy $3 \times 22.4 \ L = 67.2 \ L$.
The molar mass of $KCl = 39 + 35.5 = 74.5 \ g \ mol^{-1}$.
So,$2 \ moles$ of $KCl = 2 \times 74.5 \ g = 149 \ g$.
According to the reaction,$67.2 \ L$ of $O_2$ is produced by $149 \ g$ of $KCl$.
Therefore,$33.6 \ L$ of $O_2$ will be produced by: $\frac{149 \times 33.6}{67.2} = 74.5 \ g$ of $KCl$.

(A) The chemical equation for the hydrogenation of ethene is: $CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3$
From the stoichiometry,$1 \ mole$ of ethene ($C_2H_4$,molar mass $28 \ g/mol$) produces $1 \ mole$ of ethane ($C_2H_6$,molar mass $30 \ g/mol$).
The molar mass of ethane $(C_2H_6)$ is $(2 \times 12) + (6 \times 1) = 30 \ g/mol$.
To prepare $6.0 \ g$ of ethane,the number of moles of ethane required is $n = \frac{mass}{molar \ mass} = \frac{6.0 \ g}{30 \ g/mol} = 0.2 \ mol$.
Since $1 \ mole$ of ethene produces $1 \ mole$ of ethane,$0.2 \ mole$ of ethene is required to produce $0.2 \ mole$ of ethane.

(B) The balanced chemical equation is: $2 Na(s) + 2 C_2H_5OH(\ell) \rightarrow 2 C_2H_5O^-Na^+ + H_2(g) \uparrow$
From the stoichiometry,$2 \times 23 \ g$ of $Na$ produces $1 \text{ mole}$ of $H_2$ gas.
$46 \ g$ of $Na$ is equal to $2 \text{ moles}$ of $Na$.
Since $2 \times 23 \ g$ $(46 \ g)$ of $Na$ produces $1 \text{ mole}$ of $H_2$,the amount of $H_2$ produced is $1 \text{ mole}$.
Mass of $1 \text{ mole}$ of $H_2 = 2 \ g$.
Converting to $kg$: $2 \ g = 2 \times 10^{-3} \ kg$.

(A) The formula for atom economy is defined as:
$\text{Atom Economy} = \left( \frac{\text{Formula weight of desired product}}{\text{Sum of formula weights of all reactants}} \right) \times 100 \%$
Given:
$\text{Sum of formula weights of all reactants} = 274 \ u$
$\text{Atom Economy} = 50 \%$
Substituting the values into the formula:
$50 = \left( \frac{\text{Formula weight of desired product}}{274 \ u} \right) \times 100$
$0.5 = \frac{\text{Formula weight of desired product}}{274 \ u}$
$\text{Formula weight of desired product} = 0.5 \times 274 \ u = 137 \ u$
Therefore,the correct option is $A$.

(B) The molar mass $(M)$ of a gas is related to its vapour density $(VD)$ by the formula: $M = 2 \times VD$.
Given $VD = 16$,so $M = 2 \times 16 = 32 \ g/mol$.
The number of moles $(n)$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol$.
At $STP$,$1 \ mol$ of an ideal gas occupies $22.4 \ dm^3$.
Therefore,the volume occupied by $0.25 \ mol$ of gas is: $V = 0.25 \ mol \times 22.4 \ dm^3/mol = 5.6 \ dm^3$.

(D) At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$ volume.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Thus,$22.4 \ dm^3$ of $CO_2$ at $STP$ has a mass of $44 \ g$.
Therefore,the mass of $4.48 \ dm^3$ of $CO_2$ at $STP$ is calculated as:
$\text{Mass} = \frac{44 \ g}{22.4 \ dm^3} \times 4.48 \ dm^3 = 8.8 \ g$.
To convert the mass into $kg$,we divide by $1000$:
$8.8 \ g = 8.8 \times 10^{-3} \ kg$.

(C) The balanced chemical equation for the combustion of methane is:
$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$
According to the stoichiometry of the reaction,$1 \ mole$ of $CH_4$ requires $2 \ moles$ of $O_2$.
At $\text{S.T.P.}$,$1 \ mole$ of any gas occupies $22.4 \ dm^3$.
Therefore,$2 \ moles$ of $O_2$ occupy $2 \times 22.4 \ dm^3 = 44.8 \ dm^3$.
For $0.25 \ mole$ of $CH_4$,the volume of $O_2$ required is:
$V = 0.25 \times 44.8 \ dm^3 = 11.2 \ dm^3$.

(D) At $\text{STP}$, $1 \ mol$ of any ideal gas occupies a volume of $22.4 \ dm^3$.
Therefore, the volume occupied by $2.5 \ mol$ of ammonia gas is calculated as:
$V = n \times 22.4 \ dm^3 \ mol^{-1}$
$V = 2.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 56.0 \ dm^3$.

(B) $\% \text{ atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants}} \times 100$
$\% \text{ atom economy} = \frac{75}{225} \times 100 = 33.3\%$

(C) The balanced chemical equation for the combustion of methane is: $CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2 O$
Calculate the moles of methane $(CH_4)$: Molar mass of $CH_4 = 12 + (4 \times 1) = 16 \ g/mol$. Moles $= \frac{1.6 \ g}{16 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of $CH_4$ produces $2 \ mol$ of $H_2 O$. Therefore,$0.1 \ mol$ of $CH_4$ will produce $0.1 \times 2 = 0.2 \ mol$ of $H_2 O$.
Calculate the mass of water $(H_2 O)$: Molar mass of $H_2 O = (2 \times 1) + 16 = 18 \ g/mol$. Mass $= 0.2 \ mol \times 18 \ g/mol = 3.6 \ g$.

(A) $\text{Atom economy} = \frac{\text{Mass of desired product}}{\text{Molar mass of all reactants}} \times 100$
$\text{Atom economy} = \frac{27}{36} \times 100 = 75 \%$

(D) The balanced chemical equation for the synthesis of ammonia is:
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
According to Gay-Lussac's Law of Gaseous Volumes,gases react in simple ratios by volume at constant temperature and pressure.
From the stoichiometry of the reaction,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.
Therefore,to prepare $2 \ L$ of $NH_3$,we require $3 \ L$ of $H_2$ gas.

(B) The dehydration reaction of cane sugar (sucrose) with concentrated $H_2SO_4$ is given by:
$C_{12}H_{22}O_{11} \xrightarrow{\text{Conc. } H_2SO_4} 12C + 11H_2O$
From the stoichiometry,$1 \ mol$ of $C_{12}H_{22}O_{11}$ yields $11 \ mol$ of $H_2O$.
The molar mass of $C_{12}H_{22}O_{11} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \ g/mol$.
Number of moles of sugar $= \frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of sugar produces $11 \ mol$ of $H_2O$,then $0.1 \ mol$ of sugar produces $0.1 \times 11 = 1.1 \ mol$ of $H_2O$.
The mass of $1.1 \ mol$ of $H_2O = 1.1 \ mol \times 18 \ g/mol = 19.8 \ g$.

(A) The dehydration reaction of sugar $(C_{12}H_{22}O_{11})$ by conc. $H_2SO_4$ is given by:
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
First,calculate the number of moles of sugar:
Molar mass of $C_{12}H_{22}O_{11} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \ g/mol$.
Moles of sugar $= \frac{17.1 \ g}{342 \ g/mol} = 0.05 \ mol$.
According to the stoichiometry,$1 \ mol$ of sugar produces $12 \ mol$ of carbon (charcoal).
Therefore,$0.05 \ mol$ of sugar produces $0.05 \times 12 = 0.6 \ mol$ of carbon.
Mass of charcoal $= 0.6 \ mol \times 12 \ g/mol = 7.2 \ g$.

(A) The balanced chemical equation for the reaction is: $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$
From the stoichiometry,$1 \ \text{mole}$ of $H_2O$ $(18 \ g)$ is produced by $0.5 \ \text{moles}$ of $O_2$.
At $S.T.P$,$1 \ \text{mole}$ of gas occupies $22.4 \ dm^3$,so $0.5 \ \text{moles}$ of $O_2$ occupies $11.2 \ dm^3$.
Thus,$11.2 \ dm^3$ of $O_2$ is required to produce $18 \ g$ of $H_2O$.
For $9 \ g$ of $H_2O$,the volume of $O_2$ required is: $\frac{11.2 \ dm^3}{18 \ g} \times 9 \ g = 5.6 \ dm^3$.

(C) The balanced chemical equation for the decomposition of potassium chlorate is:
$2KClO_3 \longrightarrow 2KCl + 3O_2$
Calculate the molar mass of $KClO_3$:
$M(KClO_3) = 39 + 35.5 + 3 \times 16 = 122.5 \ g/mol$
Calculate the molar mass of $KCl$:
$M(KCl) = 39 + 35.5 = 74.5 \ g/mol$
Calculate the moles of $KClO_3$ present:
$n(KClO_3) = \frac{12.25 \ g}{122.5 \ g/mol} = 0.1 \ mol$
From the stoichiometry of the reaction,$2 \ mol$ of $KClO_3$ produces $2 \ mol$ of $KCl$. Therefore,$0.1 \ mol$ of $KClO_3$ will produce $0.1 \ mol$ of $KCl$.
Calculate the mass of $KCl$ produced:
$Mass = n \times M = 0.1 \ mol \times 74.5 \ g/mol = 7.45 \ g$

(D) The balanced chemical equation is: $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$.
According to the stoichiometry of the reaction,$2 \ moles$ of $KClO_3$ produce $3 \ moles$ of $O_2$ gas.
At $S.T.P.$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \ L \ mol^{-1}$.
Therefore,the volume of $3 \ moles$ of $O_2$ gas produced is $3 \times 22.4 \ L = 67.2 \ L$.

(C) The balanced chemical equation for the combustion of methane is:
$CH_{4}(g) + 2O_{2}(g) \longrightarrow CO_{2}(g) + 2H_{2}O(l)$
From the stoichiometry of the reaction,$1 \ mol$ of $CH_{4}$ requires $2 \ mol$ of $O_{2}$ for complete combustion.
At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ dm^{3}$.
Therefore,$2 \ mol$ of $O_{2}$ occupies $2 \times 22.4 \ dm^{3} = 44.8 \ dm^{3}$.
For $0.25 \ mol$ of $CH_{4}$,the volume of $O_{2}$ required is:
$0.25 \ mol \times (2 \ mol \ O_{2} / 1 \ mol \ CH_{4}) \times 22.4 \ dm^{3}/mol = 0.5 \ mol \times 22.4 \ dm^{3}/mol = 11.2 \ dm^{3}$.
Thus,the correct option is $C$.

(A) The balanced chemical equation for the reaction is: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
From the stoichiometry of the reaction,the molar masses are:
$N_2 = 28 \ g/mol$,$H_2 = 2 \ g/mol$,$NH_3 = 17 \ g/mol$.
For $2 \ moles$ of $NH_3$ $(2 \times 17 \ g = 34 \ g)$,the amount of $H_2$ required is $3 \ moles$ $(3 \times 2 \ g = 6 \ g)$.
Therefore,to produce $34 \ g$ of ammonia,$6 \ g$ of dihydrogen is required.

(C) $\text{Percent atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants used in the reaction}} \times 100$
$\text{Percent atom economy} = \frac{65 \ u}{78 \ u} \times 100 = 83.33 \% \approx 83 \%$

(A) The formula for atom economy is given by: $\text{Atom economy} = \frac{\text{Formula weight of desired product}}{\text{Formula weight of all reactants}} \times 100$
Given: $\text{Atom economy} = 75$,$\text{Product weight} = 54 \ u$.
Let the reactant weight be $x$.
$75 = \frac{54}{x} \times 100$
$x = \frac{54 \times 100}{75}$
$x = \frac{5400}{75} = 72 \ u$
Therefore,the formula weight of the reactant is $72 \ u$.

(A) The formula for percent atom economy is:
$\text{Percent atom economy} = \frac{\text{Formula weight of desired product}}{\text{Formula weight of all reactants}} \times 100$
Given:
$\text{Formula weight of reactants} = 45 \ u$
$\text{Formula weight of products} = 35 \ u$
Calculation:
$\text{Percent atom economy} = \frac{35}{45} \times 100 = 77.77... \% \approx 77.8 \%$

(B) $\text{Percentage atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants}} \times 100$
$\text{Percentage atom economy} = \frac{46}{92} \times 100 = 50 \%$

(C) Atom economy is a measure of the efficiency of a chemical reaction in terms of how many atoms from the reactants end up in the desired final product.
It is calculated using the formula:
$\%$ Atom economy $= \frac{\text{Formula weight of desired product}}{\text{Sum of formula weight of all reactants}} \times 100$

(B) Atom economy is the conversion efficiency of a chemical process in terms of all atoms involved and the desired products produced.
$\text{Percentage Atom Economy} = \frac{\text{Formula weight of desired product}}{\text{Sum of formula weight of all reactants}} \times 100$
$= \frac{123}{246} \times 100$
$= 50.00 \%$

(A) 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 is $2$ (since it is a dibasic acid).
Therefore, the normality of $2 \,M \,H_2SO_4$ is $2 \times 2 = 4 \,N$.
Using the dilution equation $N_1 V_1 = N_2 V_2$:
$4 \,N \times V_1 = 0.2 \,N \times 100 \,mL$.
$V_1 = \frac{0.2 \times 100}{4} \,mL$.
$V_1 = 5 \,mL$.

(C) For a solution of the same substance,the dilution formula is used:
$N_1V_1 = N_2V_2$
Here,$N_1 = 2 \ N$ and $V_1 = 0.1 \ dm^3$.
$A$ decinormal solution means $N_2 = \frac{1}{10} \ N = 0.1 \ N$.
Substituting the values:
$2 \ N \times 0.1 \ dm^3 = 0.1 \ N \times V_2$
$V_2 = \frac{2 \times 0.1}{0.1} = 2 \ dm^3$.

(D) The volume of $1 \ mol$ of any ideal gas at $STP$ is $22.4 \ dm^3$.
Given number of moles of ammonia gas $(n) = 3 \ mol$.
Volume of gas at $STP = n \times 22.4 \ dm^3 \ mol^{-1}$.
Volume $= 3 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 67.2 \ dm^3$.

(D) The number of moles of $CO_2$ dissolved is calculated as:
$n = M \times V(L)$
$n = 0.1 \times \frac{200}{1000} = 0.02 \ mol$
At $S.T.P$,$1 \ mol$ of an ideal gas occupies $22.4 \ L$.
Therefore,the volume of $0.02 \ mol$ of $CO_2$ at $S.T.P$ is:
$V = n \times 22.4 \ L/mol$
$V = 0.02 \times 22.4 \ L = 0.448 \ L$

(D) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is directly proportional to its number of moles in a mixture at constant temperature and volume $(P_i = n_i \times \frac{RT}{V})$.
For the partial pressure of $Ne$ to be equal to the partial pressure of $He$,their number of moles must be equal $(n_{Ne} = n_{He})$.
First,calculate the moles of $He$ $(n_{He})$:
$n_{He} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{4 \ g/mol} = 1 \ mol$.
Since $n_{Ne} = n_{He}$,the required moles of $Ne$ is $1 \ mol$.
Now,calculate the mass of $Ne$:
$\text{Mass of } Ne = n_{Ne} \times \text{molar mass of } Ne = 1 \ mol \times 20 \ g/mol = 20 \ g$.

(C) Moles of $N_2 = \frac{7}{28} = 0.25 \ mol$
Moles of $Ar = \frac{8}{40} = 0.20 \ mol$
Total moles $= 0.25 + 0.20 = 0.45 \ mol$
Mole fraction of $N_2$ $(X_{N_2})$ $= \frac{0.25}{0.45} = \frac{5}{9}$
Partial pressure of $N_2 = X_{N_2} \times P_{total}$
Partial pressure of $N_2 = \frac{5}{9} \times 27 \ bar = 15 \ bar$

(A) The balanced chemical equation is: $H_{2(g)} + Cl_{2(g)} \longrightarrow 2 HCl_{(g)}$.
According to Avogadro's Law,at constant temperature and pressure,the volume of gases is proportional to the number of moles.
Initial volume of reactants: $V_1 = 100 \ mL (H_2) + 100 \ mL (Cl_2) = 200 \ mL$.
Final volume of products: $V_2 = 200 \ mL (HCl)$.
Change in volume: $\Delta V = V_2 - V_1 = 200 \ mL - 200 \ mL = 0 \ mL = 0 \ dm^3$.
Work done $(W)$ is given by: $W = -P_{ext} \Delta V$.
Since $\Delta V = 0$,$W = -1 \ bar \times 0 \ dm^3 = 0 \ J$.

(A) The work done in an isothermal reversible expansion is given by the formula: $W = -2.303 nRT \log_{10} (V_2/V_1)$.
Since $R$, $T$, and the change in volume $(\Delta V = 10 \ dm^3)$ are constant, the work done $W$ is directly proportional to the number of moles $n$ $(W \propto n)$.
The number of moles $n$ is given by $n = \text{mass} / \text{molar mass}$.
Since the mass is equal for all gases, $n \propto 1/M$, where $M$ is the molar mass.
Therefore, $W \propto 1/M$.
To maximize the work done, we need the gas with the smallest molar mass.
The molar masses are: $H_2 = 2 \ g/mol$, $N_2 = 28 \ g/mol$, $Cl_2 = 71 \ g/mol$, and $O_2 = 32 \ g/mol$.
Since $H_2$ has the lowest molar mass, the work done is maximum for $H_2$.

(C) The molar mass of methane $(CH_{4})$ is $16 \ g \ mol^{-1} = 16 \times 10^{-3} \ kg \ mol^{-1}$.
Given that the heat of combustion for $1 \ mol$ $(16 \times 10^{-3} \ kg)$ of $CH_{4}$ is $-800 \ kJ$.
To find the heat of combustion for $4 \times 10^{-4} \ kg$ of $CH_{4}$,we use the unitary method:
$\Delta H = \frac{-800 \ kJ}{16 \times 10^{-3} \ kg} \times (4 \times 10^{-4} \ kg)$
$\Delta H = -800 \times \frac{4 \times 10^{-4}}{16 \times 10^{-3}}$
$\Delta H = -800 \times \frac{4}{16} \times 10^{-1}$
$\Delta H = -800 \times 0.25 \times 0.1 = -20 \ kJ$.

(C) The chemical reaction between ethanol and sodium metal is:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$
From the stoichiometry,$2 \ mol$ of ethanol produces $1 \ mol$ of $H_2$ gas.
At $S.T.P.$,$1 \ mol$ of any gas occupies $22400 \ mL$.
Therefore,$22400 \ mL$ of $H_2$ is produced by $2 \times 46 \ g$ of ethanol.
$280 \ mL$ of $H_2$ is produced by:
$\frac{2 \times 46 \times 280}{22400} \ g$
$= \frac{92 \times 280}{22400} \ g$
$= \frac{92}{80} \ g = 1.15 \ g$.
Thus,the correct option is $C$.

(B) The reaction of a metal '$M$' with an alcohol $(ROH)$ can be represented as:
$M + nROH \rightarrow M(OR)_n + \frac{n}{2} H_2$
Given that $1 \ mol$ of '$M$' produces $1.5 \ mol$ of $H_2$,we can equate the stoichiometric coefficient of $H_2$ to $1.5$:
$\frac{n}{2} = 1.5$
$n = 1.5 \times 2 = 3$
Therefore,the valency of the metal '$M$' is $3$.

(D) The balanced chemical equation for the reaction of ethanol with sodium metal is:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$
From the stoichiometry,$2 \ mol$ of ethanol produces $1 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.
Given volume of $H_2 = 560 \ mL = 0.56 \ L$.
Moles of $H_2 = \frac{560 \ mL}{22400 \ mL \ mol^{-1}} = 0.025 \ mol$.
According to the equation,moles of ethanol required = $2 \times \text{moles of } H_2 = 2 \times 0.025 = 0.05 \ mol$.
Mass of ethanol = $\text{moles} \times \text{molar mass} = 0.05 \ mol \times 46 \ g \ mol^{-1} = 2.3 \ g$.
Therefore,the correct option is $D$.

(B) The balanced chemical equation for the reaction is: $(COOH)_2 + 2NaOH \rightarrow (COONa)_2 + 2H_2O$.
From the stoichiometry,$1 \ mole$ of oxalic acid reacts with $2 \ moles$ of $NaOH$.
Number of moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{0.064 \ g}{40 \ g/mol} = 0.0016 \ mol$.
Since $2 \ moles$ of $NaOH$ react with $1 \ mole$ of oxalic acid,the moles of oxalic acid $= \frac{0.0016}{2} = 0.0008 \ mol$.
Volume of oxalic acid $= 25 \ cm^3 = 0.025 \ L$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.0008 \ mol}{0.025 \ L} = 0.032 \ M$.

(B) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2$.
From the stoichiometry,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } H_2C_2O_4$.
Using the equivalence concept: $n_{factor} \times M \times V = \text{constant}$.
For $KMnO_4$ in acidic medium,$n_{factor} = 5$.
For $H_2C_2O_4$,$n_{factor} = 2$.
Equating equivalents: $n_{factor,1} \times M_1 \times V_1 = n_{factor,2} \times M_2 \times V_2$.
$5 \times 0.025 \times 20 = 2 \times 0.1 \times V$.
$2.5 = 0.2 \times V$.
$V = \frac{2.5}{0.2} = 12.5 \ mL$.

(C) Let the mass of $CaO$ be $x \text{ g}$. Then the mass of $BaO$ is $(10 - x) \text{ g}$.
Moles of $HCl = \text{Molarity} \times \text{Volume in L} = 2.5 \times 0.1 = 0.25 \text{ mol}$.
The reactions are:
$CaO + 2HCl \rightarrow CaCl_2 + H_2O$
$BaO + 2HCl \rightarrow BaCl_2 + H_2O$
Total moles of $HCl = 2 \times (\text{moles of } CaO) + 2 \times (\text{moles of } BaO)$
$0.25 = 2 \times (\frac{x}{56} + \frac{10 - x}{153})$
$0.125 = \frac{153x + 560 - 56x}{56 \times 153}$
$0.125 \times 8568 = 97x + 560$
$1071 = 97x + 560$
$97x = 511$
$x = 5.268 \text{ g}$
Percentage of $CaO = \frac{5.268}{10} \times 100 = 52.68 \% \approx 52.6 \%$.

(A) Orthophosphoric acid $(H_{3}PO_{4})$ is a tribasic acid,meaning its basicity is $3$.
The relationship between normality and molarity is given by the formula: $\text{Normality} = \text{Molarity} \times \text{basicity}$.
Substituting the given values: $\text{Normality} = 3 \ M \times 3 = 9 \ N$.

(A) Let the total number of oxide ions be $100$. Then the number of metal ions $M$ is $96$.
Let the number of $M^{2+}$ ions be $x$ and the number of $M^{3+}$ ions be $(96 - x)$.
Since the compound is electrically neutral,the total positive charge must equal the total negative charge.
Total negative charge $= 100 \times 2 = 200$.
Total positive charge $= 2x + 3(96 - x) = 200$.
$2x + 288 - 3x = 200$.
$-x = 200 - 288 = -88$.
$x = 88$ (number of $M^{2+}$ ions).
Number of $M^{3+}$ ions $= 96 - 88 = 8$.
Percentage of $M^{3+} = \frac{8}{96} \times 100 = 8.33 \% \approx 8.3 \%$.

(C) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$
Initially: $20 \ mL$ of $CH_4$ and $50 \ mL$ of $O_2$ are taken.
According to the stoichiometry,$1 \ volume$ of $CH_4$ requires $2 \ volumes$ of $O_2$.
So,$20 \ mL$ of $CH_4$ will react with $2 \times 20 = 40 \ mL$ of $O_2$.
Remaining $O_2 = 50 \ mL - 40 \ mL = 10 \ mL$.
Volume of $CO_2$ produced = $20 \ mL$.
Since $H_2O$ is produced as a liquid at room temperature,its volume is negligible.
Total volume of gas left = $Volume \ of \ remaining \ O_2 + Volume \ of \ CO_2 = 10 \ mL + 20 \ mL = 30 \ mL$.

(C) The balanced chemical equation is: $Fe_{2}O_{3} + 3 CO \longrightarrow 2 Fe + 3 CO_{2}$.
According to the stoichiometry of the reaction,$1 \ mole$ of $Fe_{2}O_{3}$ reacts with $3 \ moles$ of $CO$.
At $STP$,the volume of $1 \ mole$ of any ideal gas is $22.4 \ dm^{3}$.
Therefore,the volume of $3 \ moles$ of $CO = 3 \times 22.4 \ dm^{3} = 67.2 \ dm^{3}$.

(A) The chemical reactions involved are:
$CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaCl$
$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$
From the stoichiometry,$1 \ mol$ of $CaO$ is obtained from $1 \ mol$ of $CaCO_3$,which in turn comes from $1 \ mol$ of $CaCl_2$.
Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.
Molar mass of $CaCl_2 = 40 + 2 \times 35.5 = 111 \ g/mol$.
Given mass of $CaO = 0.56 \ g$,which is $0.56 / 56 = 0.01 \ mol$.
Therefore,moles of $CaCl_2 = 0.01 \ mol$.
Mass of $CaCl_2 = 0.01 \times 111 = 1.11 \ g$.
Mass of $NaCl$ in the mixture = Total mass - Mass of $CaCl_2 = 4.44 \ g - 1.11 \ g = 3.33 \ g$.
Percentage of $NaCl = (3.33 / 4.44) \times 100 = 75\%$.

(A) The chemical reactions involved are:
$CaCl_{2} + Na_{2}CO_{3} \rightarrow CaCO_{3} + 2NaCl$
$CaCO_{3} \xrightarrow{\Delta} CaO + CO_{2}$
From the stoichiometry,$1 \ mol$ of $CaO$ is obtained from $1 \ mol$ of $CaCO_{3}$,which in turn comes from $1 \ mol$ of $CaCl_{2}$.
Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.
Moles of $CaO = \frac{0.56 \ g}{56 \ g/mol} = 0.01 \ mol$.
Therefore,moles of $CaCl_{2} = 0.01 \ mol$.
Molar mass of $CaCl_{2} = 40 + (2 \times 35.5) = 111 \ g/mol$.
Mass of $CaCl_{2} = 0.01 \ mol \times 111 \ g/mol = 1.11 \ g$.
Mass of $NaCl$ in the mixture $= 4.44 \ g - 1.11 \ g = 3.33 \ g$.
Percentage of $NaCl = \frac{3.33 \ g}{4.44 \ g} \times 100 = 75 \%$.