Chemical stoichiometry Questions in English

(C) The thermal decomposition of silver carbonate is given by the equation: $2Ag_2CO_3 \xrightarrow{\Delta} 4Ag + 2CO_2 + O_2$.
The molar mass of $Ag_2CO_3$ is $(2 \times 108) + 12 + (3 \times 16) = 276 \, g/mol$.
The molar mass of $Ag$ is $108 \, g/mol$.
From the stoichiometry,$2 \, mol$ of $Ag_2CO_3$ $(2 \times 276 = 552 \, g)$ produces $4 \, mol$ of $Ag$ $(4 \times 108 = 432 \, g)$.
The residue obtained is metallic silver $(Ag)$.
For $2.76 \, g$ of $Ag_2CO_3$,the weight of $Ag$ produced is $\frac{432 \times 2.76}{552} = 2.16 \, g$.

(A) The neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
According to the law of equivalence,$N_1V_1 = N_2V_2$.
For $NaOH$,$Molarity = Normality$ because its $n$-factor is $1$. So,$N_{NaOH} = 0.164 \ N$.
Using the formula $N_{acid} \times V_{acid} = N_{base} \times V_{base}$:
$N_{acid} \times 25 \ mL = 0.164 \ N \times 32.63 \ mL$.
$N_{acid} = \frac{0.164 \times 32.63}{25} = 0.214 \ N$.
Since $Molarity = \frac{Normality}{n\text{-factor}}$ and the $n$-factor (basicity) of $H_2SO_4$ is $2$:
$Molarity = \frac{0.214}{2} = 0.107 \ M$.

(C) The neutralization reaction is: $NaOH + HCl \rightarrow NaCl + H_2O$.
According to the law of equivalence,the number of moles of $H^+$ must equal the number of moles of $OH^-$.
For $HCl$,the number of moles $= Molarity \times Volume = 0.4 \, M \times 30 \, cm^3 = 12 \, mmol$.
Since $NaOH$ is a monobasic base,$1 \, mol$ of $NaOH$ reacts with $1 \, mol$ of $HCl$.
Therefore,$12 \, mmol$ of $NaOH$ is required.
$Volume \, of \, NaOH = \frac{Moles}{Molarity} = \frac{12 \, mmol}{0.6 \, M} = 20 \, cm^3$.

(C) The chemical reaction for the conversion of sodium dihydrogen phosphate to trisodium phosphate is:
$NaH_2PO_4 + 2NaOH \rightarrow Na_3PO_4 + 2H_2O$
From the stoichiometry,$1 \text{ mole}$ of $NaH_2PO_4$ requires $2 \text{ moles}$ of $NaOH$.
Number of moles of $NaH_2PO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \ g}{120 \ g/mol} = 0.1 \text{ mol}$.
Moles of $NaOH$ required = $2 \times 0.1 = 0.2 \text{ mol}$.
Using the molarity formula $M = \frac{n}{V(L)}$,we have $V(L) = \frac{n}{M} = \frac{0.2 \text{ mol}}{1 \text{ M}} = 0.2 \text{ L}$.
Converting to $mL$,$V = 0.2 \times 1000 = 200 \ mL$.

(A) The reaction is: $H_2O_2 + 2KI + 2H^+ \rightarrow 2H_2O + I_2$ and $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.
Equivalents of $H_2O_2$ = Equivalents of $I_2$ = Equivalents of $Na_2S_2O_3$.
Normality of $Na_2S_2O_3$ = $Molarity \times n\text{-factor} = 0.3 \times 1 = 0.3 \, N$.
Equivalents of $Na_2S_2O_3 = 0.3 \times 20 \times 10^{-3} = 6 \times 10^{-3}$.
Normality of $H_2O_2$ in $25 \, mL$ solution = $\frac{6 \times 10^{-3}}{25 \times 10^{-3}} = 0.24 \, N$.
Volume strength of $H_2O_2 = 5.6 \times \text{Normality} = 5.6 \times 0.24 = 1.344 \, mL$.

(A) The chemical reaction is: $CH_3-CH=CH_2 + Br_2 \rightarrow CH_3-CH(Br)-CH_2(Br)$
The molar mass of propene $(C_3H_6)$ is $(3 \times 12) + (6 \times 1) = 42\, g/mol$.
The molar mass of bromine $(Br_2)$ is $2 \times 80 = 160\, g/mol$.
According to the stoichiometry,$1$ mole of propene reacts with $1$ mole of bromine.
Thus,$42\, g$ of propene reacts with $160\, g$ of bromine.
Therefore,$21\, g$ of propene will react with $\frac{160}{42} \times 21 = 80\, g$ of bromine.

(D) The stoichiometry of the reactions is as follows:
$1 \ mol$ of $CaC_2$ produces $1 \ mol$ of $C_2H_2$ (acetylene).
$1 \ mol$ of $C_2H_2$ produces $1 \ mol$ of $C_2H_4$ (ethylene).
$n \ mol$ of $C_2H_4$ produces $1 \ mol$ of polyethylene $(-CH_2-CH_2-)_n$.
Thus,$1 \ mol$ of $CaC_2$ produces $1 \ mol$ of ethylene units in polyethylene.
Molar mass of $CaC_2 = 40 + 2 \times 12 = 64 \ g/mol$.
Molar mass of ethylene unit $(C_2H_4) = 2 \times 12 + 4 \times 1 = 28 \ g/mol$.
Since $64 \ g$ of $CaC_2$ produces $28 \ g$ of polyethylene,$64.1 \ kg$ of $CaC_2$ will produce approximately $28.04 \ kg$ of polyethylene.
Rounding to the nearest given option,the answer is $28 \ kg$.

(D) The molarity formula is given by $M = \frac{w}{M_m \times V(L)}$,where $w$ is the mass in grams,$M_m$ is the molar mass,and $V$ is the volume in liters.
For $Na_2CO_3$,the molar mass $M_m = (2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$.
Given $M = 0.25 \ M$ and $V = 250 \ mL = 0.25 \ L$.
Substituting the values: $0.25 = \frac{w}{106 \times 0.25}$.
$w = 0.25 \times 106 \times 0.25 = 6.625 \ g$.
Rounding to two decimal places,we get $6.63 \ g$.

(D) The molarity $(M)$ of $H_2SO_4$ is $1 \ M$. The n-factor for $H_2SO_4$ is $2$.
Normality $(N)$ = Molarity $(M)$ $\times$ n-factor = $1 \times 2 = 2 \ N$.
Initial volume $(V_1)$ = $1 \ L$.
Final volume $(V_2)$ = $1 \ L + 5 \ L = 6 \ L$.
Using the dilution formula $N_1V_1 = N_2V_2$:
$2 \times 1 = N_2 \times 6$
$N_2 = \frac{2}{6} = 0.33 \ N$.

(D) The formula for normality is $\text{Normality} = \text{Molarity} \times \text{n-factor}$.
For phosphorous acid $(H_3PO_3)$,the structure contains two $P-OH$ bonds,making it a dibasic acid.
Therefore,the n-factor (basicity) of $H_3PO_3$ is $2$.
$\text{Normality} = 0.3 \ M \times 2 = 0.6 \ N$.

(B) The neutralization reaction is $NaOH + HCl \rightarrow NaCl + H_2O$.
For neutralization,the number of gram equivalents of $NaOH$ must be equal to the number of gram equivalents of $HCl$.
Number of gram equivalents of $HCl = N \times V \text{ (in Liters)} = 0.1 \times \frac{1500}{1000} = 0.15 \text{ equivalents}$.
Since the equivalent weight of $NaOH$ is $40 \ g/eq$,the mass of $NaOH$ required is $0.15 \times 40 = 6 \ g$.

(B) The relationship between normality $(N)$ and molarity $(M)$ is given by the formula: $N = M \times \text{n-factor}$.
For $H_2SO_4$,the n-factor (basicity) is $2$ because it provides $2 \ H^+$ ions per molecule.
Given molarity $M = 2.3 \ M$.
Therefore,$N = 2.3 \times 2 = 4.6 \ N$.

(C) The normality of the final solution is calculated using the dilution formula: $N_1V_1 = N_2V_2$.
Here,$N_1 = 36 \ N$,$V_1 = 50 \ mL$,and $V_2 = 50 \ mL + 50 \ mL = 100 \ mL$.
$36 \times 50 = N_2 \times 100$.
$N_2 = \frac{36 \times 50}{100} = 18 \ N$.
For $H_2SO_4$,the n-factor is $2$. The relationship between normality and molarity is $N = M \times \text{n-factor}$.
$18 = M \times 2$,so $M = 9 \ M$.

(C) The molar mass of urea $(NH_2CONH_2)$ is $60 \, g/mol$.
The number of moles of urea $= \frac{\text{mass}}{\text{molar mass}} = \frac{240 \, g}{60 \, g/mol} = 4 \, mol$.
Active mass is defined as the molar concentration (molarity) of the substance.
Active mass $= \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{4 \, mol}{10 \, L} = 0.4 \, mol/L$.

(D) The formula for the normality of a mixture is $N_{mix}V_{mix} = N_1V_1 + N_2V_2 + N_3V_3$.
Given:
$N_1 = 1 \ N, V_1 = 5 \ mL$
$N_2 = 1/2 \ N, V_2 = 20 \ mL$
$N_3 = 1/3 \ N, V_3 = 30 \ mL$
$V_{mix} = 1 \ L = 1000 \ mL$
Substituting the values:
$N_{mix} \times 1000 = (1 \times 5) + (1/2 \times 20) + (1/3 \times 30)$
$N_{mix} \times 1000 = 5 + 10 + 10$
$N_{mix} \times 1000 = 25$
$N_{mix} = 25 / 1000 = 1 / 40 \ N$
Thus,the normality is $N/40$.

(A) The molarity $(M)$ is defined as the number of moles of solute per liter of solution.
The formula is $M = \frac{w}{m \times V(L)}$,where $w$ is the mass in grams,$m$ is the molar mass of $HCl$ $(36.5 \ g/mol)$,and $V(L)$ is the volume in liters.
Given: $M = 0.52 \ M$,$V = 150 \ mL = 0.15 \ L$,and $m = 36.5 \ g/mol$.
Substituting the values: $0.52 = \frac{w}{36.5 \times 0.15}$.
Calculating $w$: $w = 0.52 \times 36.5 \times 0.15 = 2.847 \ g \approx 2.84 \ g$.

(A) The formula for normality $(N)$ is $N = \frac{\text{mass in } g \times 1000}{\text{Equivalent mass} \times \text{Volume in } mL}$.
First,calculate the equivalent mass $(E)$ of $H_3PO_4$. The molar mass of $H_3PO_4$ is $98 \ g/mol$ and its basicity is $3$.
$E = \frac{98}{3} \approx 32.67 \ g/eq$.
Now,substitute the values into the formula:
$N = \frac{4.9 \times 1000}{32.67 \times 500} = \frac{4900}{16335} \approx 0.3 \ N$.

(B) The molarity formula is given by $M = \frac{n}{V(L)}$,where $n$ is the number of moles and $V(L)$ is the volume in liters.
Given: $M = 0.8 \ M$ and $n = 0.1 \ mole$.
Substituting the values: $0.8 = \frac{0.1}{V(L)}$.
$V(L) = \frac{0.1}{0.8} = 0.125 \ L$.
To convert liters to milliliters,multiply by $1000$: $0.125 \ L \times 1000 \ mL/L = 125 \ mL$.

(D) The relationship between normality and molarity is given by the formula: $\text{Normality} = \text{Molarity} \times \text{Basicity}$.
For $H_2SO_4$,the basicity is $2$ because it is a dibasic acid.
Given that the normality is $0.2 \ N$,we have: $0.2 = \text{Molarity} \times 2$.
Therefore,$\text{Molarity} = \frac{0.2}{2} = 0.1 \ M$.

(D) The molar mass of hydrogen gas $(H_2)$ is $2 \ g/mol$.
Number of moles of $H_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{20 \ g}{2 \ g/mol} = 10 \ mol$.
Molar concentration (Molarity) is defined as the number of moles of solute per liter of solution.
$\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{10 \ mol}{5 \ L} = 2 \ M$.
Therefore,the correct option is $(D)$.

(A) Given: Molecular weight $(M_w)$ = $200 \ g/mol$.
Since the acid is dibasic,its basicity is $2$.
Equivalent weight $(E)$ = $\frac{M_w}{\text{basicity}} = \frac{200}{2} = 100 \ g/eq$.
Strength $(N)$ = $0.1 \ N$ (decinormal).
Volume $(V)$ = $100 \ mL = 0.1 \ L$.
Using the formula: $N = \frac{w}{E \times V(L)}$,where $w$ is the weight in grams.
$0.1 = \frac{w}{100 \times 0.1}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1 \ g$.

(A) The formula for normality is $N = \frac{w}{E \times V_{(L)}}$,where $w$ is the weight in grams,$E$ is the equivalent weight,and $V_{(L)}$ is the volume in liters.
For $NaOH$,the equivalent weight $E = 40 \ g/eq$.
The volume $V = 250 \ cm^3 = 0.25 \ L$.
Substituting the values: $0.1 = \frac{w}{40 \times 0.25}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1.0 \ g$.

(A) For neutralization,the number of equivalents of base equals the number of equivalents of acid: $N_1V_1 = N_2V_2$.
Given: $N_1 = 0.4 \ N$,$V_1 = 20 \ mL$,$V_2 = 40 \ mL$.
$0.4 \times 20 = N_2 \times 40$.
$N_2 = \frac{8}{40} = 0.2 \ N$.
For a dibasic acid,$\text{Normality} = \text{Molarity} \times \text{Basicity}$.
$0.2 = M \times 2$.
$M = 0.1 \ M$.

(C) The reaction between $CaCO_3$ and $HCl$ is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$.
At the equivalence point,the number of milliequivalents $(M.eq.)$ of $HCl$ is equal to the number of milliequivalents of $CaCO_3$.
$M.eq. \text{ of } HCl = M.eq. \text{ of } CaCO_3$.
$N \times V(mL) = \frac{\text{mass}}{\text{equivalent mass}} \times 1000$.
The equivalent mass of $CaCO_3$ is $\frac{100}{2} = 50 \ g/eq$.
$N \times 50 = \frac{1.0}{50} \times 1000$.
$N = \frac{1000}{50 \times 50} = \frac{1000}{2500} = 0.4 \ N$.

(B) The molarity of $H_2SO_4$ is $0.5 \ M$.
The normality of $H_2SO_4$ is calculated as $N = M \times \text{n-factor}$. For $H_2SO_4$,the n-factor is $2$.
Initial normality $(N_1) = 0.5 \times 2 = 1 \ N$.
Using the dilution formula $N_1V_1 = N_2V_2$:
$1 \times 1 \ L = N_2 \times 10 \ L$.
$N_2 = \frac{1}{10} = 0.1 \ N$.

(A) Using the dilution formula $N_1V_1 = N_2V_2$:
Given $N_1 = 0.5 \, N$,$V_1 = 100 \, cm^3$,and $N_2 = 0.1 \, N$ (decinormal).
Substituting the values: $0.5 \times 100 = 0.1 \times V_2$.
$V_2 = \frac{0.5 \times 100}{0.1} = 500 \, cm^3$.
The total volume $V_2$ is $500 \, cm^3$.
The volume of water to be added = $V_2 - V_1 = 500 \, cm^3 - 100 \, cm^3 = 400 \, cm^3$.

(B) Using the dilution formula: $N_1V_1 = N_2V_2$
Here,$N_1 = 10 \ N$,$V_1 = 10 \ mL$,$N_2 = 0.1 \ N$,and $V_2 = V_1 + V_{\text{water}}$.
Substituting the values: $10 \times 10 = 0.1 \times (10 + V_{\text{water}})$
$100 = 0.1 \times (10 + V_{\text{water}})$
$1000 = 10 + V_{\text{water}}$
$V_{\text{water}} = 1000 - 10 = 990 \ mL$.

(D) The equivalent weight of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $63 \ g/eq$ (Molar mass = $126 \ g/mol$,n-factor = $2$).
Normality $(N)$ is defined as the number of gram equivalents per liter of solution.
Formula: $Weight (g) = \frac{N \times V(mL) \times Eq. Wt.}{1000}$.
Substituting the values: $Weight = \frac{0.2 \times 500 \times 63}{1000}$.
$Weight = 0.1 \times 63 = 6.3 \ g$.

(A) Using the dilution formula: $M_1V_1 = M_2V_2$
Given $M_1 = 5 \, M$,$V_1 = 1 \, L$,$V_2 = 10 \, L$.
$5 \times 1 = M_2 \times 10 \Rightarrow M_2 = 0.5 \, M$.
Normality $(N)$ is related to Molarity $(M)$ by the formula: $N = M \times \text{n-factor}$.
For $H_2SO_4$,the basicity (n-factor) is $2$.
Therefore,$N = 0.5 \times 2 = 1.0 \, N$.

(A) The molarity $(M)$ is calculated using the formula: $M = \frac{w \times 1000}{m \times V \text{ (in } mL)}$.
Here,the mass of $NaOH$ $(w)$ is $1 \ g$,the molar mass of $NaOH$ $(m)$ is $40 \ g/mol$,and the volume of the solution $(V)$ is $250 \ mL$.
Substituting the values: $M = \frac{1 \times 1000}{40 \times 250} = \frac{1000}{10000} = 0.1 \ M$.
Therefore,the correct option is $A$.

(B) The number of moles of $HCl$ is calculated as: $n = \frac{1.2046 \times 10^{24}}{6.023 \times 10^{23}} = 2 \ mol$.
Since the volume of the solution is $1 \ dm^3$ $(1 \ L)$,the molarity $(M)$ is $2 \ mol/L$.
For $HCl$,the n-factor (basicity) is $1$.
Therefore,the normality $(N)$ is given by: $N = M \times n\text{-factor} = 2 \times 1 = 2 \ N$.

(B) The relationship between Normality $(N)$ and Molarity $(M)$ is given by the formula: $N = M \times \text{basicity}$.
For sulphuric acid $(H_2SO_4)$,the basicity is $2$ because it can provide $2$ replaceable $H^+$ ions.
Given $M = 2 \, M$.
Therefore,$N = 2 \times 2 = 4 \, N$.

(A) $H_3PO_3$ is a dibasic acid,meaning it provides $2$ moles of $H^+$ ions per mole of acid.
Using the principle of equivalence: $n_{acid} \times \text{basicity} = n_{base} \times \text{acidity}$.
$M_1 \times V_1 \times \text{basicity} = M_2 \times V_2 \times \text{acidity}$.
$0.1 \times 20 \times 2 = 0.1 \times V_2 \times 1$.
$4 = 0.1 \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.

(B) The neutralization reaction is: $C_6H_5COOH + NaOH \to C_6H_5COONa + H_2O$.
The molar mass of benzoic acid $(C_6H_5COOH)$ is $122 \ g/mol$.
The molar mass of $NaOH$ is $40 \ g/mol$.
Since the stoichiometry is $1:1$,the moles of $NaOH$ required equal the moles of benzoic acid.
Moles of benzoic acid = $\frac{12.2 \ g}{122 \ g/mol} = 0.1 \ mol$.
Moles of $NaOH$ required = $0.1 \ mol$.
Mass of $NaOH$ = $0.1 \ mol \times 40 \ g/mol = 4.0 \ g$.

(C) For $H_2SO_4$,the relation between Molarity $(M)$ and Normality $(N)$ is $N = M \times \text{basicity}$.
Since the basicity of $H_2SO_4$ is $2$,the normality of the concentrated acid is $N_1 = 18 \ M \times 2 = 36 \ N$.
Using the dilution formula $N_1 V_1 = N_2 V_2$:
$36 \ N \times 10 \ mL = N_2 \times 1000 \ mL$.
$N_2 = (36 \times 10) / 1000 = 0.36 \ N$.

(B) The concentration of $CaCl_2$ solution is $0.5 \, mol/L$.
$500 \, mL$ is equal to $0.500 \, L$.
The number of moles of $CaCl_2$ in $500 \, mL$ is calculated as: $0.500 \, L \times 0.5 \, mol/L = 0.25 \, moles$ of $CaCl_2$.
Since $1 \, mole$ of $CaCl_2$ dissociates to give $2 \, moles$ of chloride ions $(Cl^-)$,the moles of chloride ions will be: $2 \times 0.25 = 0.50 \, moles$.

(C) Given: $98 \ \%$ $H_2SO_4$ by weight means $98 \ g$ of $H_2SO_4$ is present in $100 \ g$ of the solution.
Step $1$: Calculate the volume of the solution.
$Volume = \frac{Mass}{Density} = \frac{100 \ g}{1.84 \ g/cc} \approx 54.35 \ cc = 0.05435 \ L$.
Step $2$: Calculate the moles of solute $(H_2SO_4)$.
Molar mass of $H_2SO_4 = 2(1) + 32 + 4(16) = 98 \ g/mol$.
Moles of $H_2SO_4 = \frac{98 \ g}{98 \ g/mol} = 1 \ mol$.
Step $3$: Calculate Molarity $(M)$.
$M = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{1 \ mol}{0.05435 \ L} \approx 18.4 \ M$.

(B) The molarity $(M)$ of a solution is defined as the number of moles of solute per liter of solution.
For sulphuric acid $(H_2SO_4)$,the n-factor (basicity) is $2$ because it releases $2$ $H^+$ ions per molecule.
The relationship between normality $(N)$ and molarity $(M)$ is given by: $N = M \times \text{n-factor}$.
For a $1\,M$ solution of $H_2SO_4$,$N = 1 \times 2 = 2\,N$.
Therefore,the molar solution of sulphuric acid is equal to $2\,N$ solution.

(A) The formula for normality is $N = \frac{w \times 1000}{Eq. \ wt. \times V(mL)}$.
For sodium carbonate $(Na_2CO_3)$,the molar mass is $106 \ g/mol$ and the n-factor is $2$ (since it provides $2 \ Na^+$ ions).
Therefore,$Eq. \ wt. = \frac{106}{2} = 53 \ g/eq$.
$A$ semi-normal solution means $N = 0.5 \ N$.
Given $V = 500 \ mL$,we substitute these values into the formula:
$0.5 = \frac{w \times 1000}{53 \times 500}$.
$w = \frac{0.5 \times 53 \times 500}{1000} = 13.25 \ g$.

(A) Using the dilution formula: $N_1 V_1 = N_2 V_2$
Given: $N_1 = 10 \, N$,$V_1 = 10 \, mL$,$N_2 = 0.1 \, N$.
Substituting the values: $10 \times 10 = 0.1 \times V_2$
$V_2 = \frac{100}{0.1} = 1000 \, mL$ (Total final volume).
Volume of water to be added $= V_2 - V_1 = 1000 \, mL - 10 \, mL = 990 \, mL$.

(C) The chemical reaction is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$.
Given mass of $CaCO_3 = 1.0 \ g$.
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Equivalent weight of $CaCO_3 = \frac{\text{Molar mass}}{n-factor} = \frac{100}{2} = 50$.
Using the law of equivalence,the number of equivalents of $HCl$ must equal the number of equivalents of $CaCO_3$.
$N \times V(L) = \frac{\text{mass}}{\text{Equivalent weight}}$.
$0.1 \times V(L) = \frac{1.0}{50}$.
$V(L) = \frac{1}{50 \times 0.1} = \frac{1}{5} = 0.2 \ L$.
Converting to $cm^3$ (or $mL$): $0.2 \ L \times 1000 \ cm^3/L = 200 \ cm^3$.

(C) The molarity $(M)$ is defined as the number of moles of solute per litre of solution.
Given mass of $NaOH$ $(w)$ = $4.0 \,g$.
Molar mass of $NaOH$ $(M_{wt})$ = $23 + 16 + 1 = 40 \,g/mol$.
Volume of solution $(V)$ = $1 \, \text{decilitre} = 0.1 \,L$.
Using the formula: $M = \frac{w}{M_{wt} \times V(L)}$.
$M = \frac{4.0}{40 \times 0.1} = \frac{4.0}{4.0} = 1 \,M$.

(A) The balanced chemical equation is: $Na_2CO_3 + H_2SO_4 \to Na_2SO_4 + CO_2 + H_2O$
Moles of $H_2SO_4 = M \times V(L) = 0.1 \times 0.1 = 0.01 \ mol$
According to the stoichiometry,$1 \ mol$ of $H_2SO_4$ produces $1 \ mol$ of $CO_2$.
Therefore,$0.01 \ mol$ of $H_2SO_4$ will produce $0.01 \ mol$ of $CO_2$.
Volume of $CO_2$ at $STP = 0.01 \ mol \times 22.4 \ L/mol = 0.224 \ L$.

(B) From the balanced chemical equation,$2 \ \text{moles}$ of $MnO_4^-$ react with $5 \ \text{moles}$ of $C_2O_4^{2-}$.
The number of moles of $KMnO_4$ used is $n = M \times V = 0.1 \ \text{mol/L} \times 0.020 \ \text{L} = 0.002 \ \text{moles}$.
According to the stoichiometry,$2 \ \text{moles}$ of $KMnO_4$ require $5 \ \text{moles}$ of oxalic acid $(C_2H_2O_4)$.
Therefore,$0.002 \ \text{moles}$ of $KMnO_4$ require $(5/2) \times 0.002 = 0.005 \ \text{moles}$ of $C_2H_2O_4$.
Checking the options:
For option $B$: $n = 0.1 \ \text{M} \times 0.050 \ \text{L} = 0.005 \ \text{moles}$.
Thus,$50 \ \text{mL}$ of $0.1 \ \text{M}$ $C_2H_2O_4$ is equivalent.

(C) The chemical equation for the reaction is: $CH_3OH + Na \to CH_3ONa + \frac{1}{2}H_2$.
The molar mass of $Na$ is $23 \ g/mol$.
Therefore,$23 \ g$ of $Na$ is equal to $1 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $Na$ produces $\frac{1}{2} \ mol$ of $H_2$ gas.

(D) The balanced chemical equation for the combustion of benzene in the gaseous state is:
$C_6H_{6(g)} + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$
The change in the number of moles of gaseous species is calculated as:
$\Delta n_{(g)} = \sum n_{p(g)} - \sum n_{r(g)}$
$\Delta n_{(g)} = (6 + 3) - (1 + 7.5)$
$\Delta n_{(g)} = 9 - 8.5 = 0.5$ or $\frac{1}{2}$