Chemical stoichiometry Questions in English

(B) At $STP$,$22.4 \, dm^3$ of $CO_2$ corresponds to $1 \, mol$ $(44 \, g)$.
So,$11.2 \, dm^3$ of $CO_2$ corresponds to $0.5 \, mol$ $(22 \, g)$.
The reaction for neutralization is: $KOH + CO_2 \rightarrow KHCO_3$.
According to the stoichiometry,$1 \, mol$ of $KOH$ $(56 \, g)$ reacts with $1 \, mol$ of $CO_2$ $(44 \, g)$.
Therefore,for $0.5 \, mol$ of $CO_2$,the required mass of $KOH = 0.5 \times 56 \, g = 28 \, g$.

(C) The chemical equations for the reactions are:
$(1) \ 2KClO_3 \rightarrow 2KCl + 3O_2$
$(2) \ 2CO + O_2 \rightarrow 2CO_2$
From equation $(2)$,$2 \ mol$ of $CO$ requires $1 \ mol$ of $O_2$ ($22.4 \ L$ at $N.T.P$).
Therefore,$84 \ L$ of $CO$ requires $\frac{84}{2} = 42 \ L$ of $O_2$.
From equation $(1)$,$3 \times 22.4 \ L$ of $O_2$ is produced by $2 \ mol$ of $KClO_3$.
So,$42 \ L$ of $O_2$ is produced by $\frac{2 \times 42}{3 \times 22.4} = 1.25 \ mol$ of $KClO_3$.
The molar mass of $KClO_3 = 39.1 + 35.5 + 3 \times 16 = 122.6 \ g/mol$.
Weight of $KClO_3 = 1.25 \ mol \times 122.6 \ g/mol = 153.25 \ g$.
Given the options,the closest value is $153.12 \ g$.

(D) The balanced chemical equation for the reduction of $Fe_2O_3$ is:
$Fe_2O_3 + 3H_2 \rightarrow 2Fe + 3H_2O$
The molar mass of $Fe_2O_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \ g/mol$.
$1 \ kg$ of $Fe_2O_3 = 1000 \ g$.
Number of moles of $Fe_2O_3 = \frac{1000 \ g}{160 \ g/mol} = 6.25 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of $Fe_2O_3$ produces $2 \ mol$ of $Fe$.
Therefore,$6.25 \ mol$ of $Fe_2O_3$ produces $6.25 \times 2 = 12.5 \ mol$ of $Fe$.
Mass of $Fe = \text{moles} \times \text{atomic mass} = 12.5 \ mol \times 56 \ g/mol = 700 \ g$.

(A) The molar mass of $N_2$ is $28 \, g/mol$.
At $NTP$,the volume of $1$ mole of any gas is $22.4 \, L$.
Density $(d)$ is defined as mass $(m)$ divided by volume $(v)$.
$d = \frac{m}{v} = \frac{28 \, g}{22.4 \, L} = 1.25 \, g/L$.

(A) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Formula: $M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$
Given: $M = 3 \ M$,Volume = $1000 \ mL = 1 \ L$.
Substituting the values: $3 = \frac{\text{moles}}{1 \ L}$.
Therefore,moles of $KCl = 3 \ mol$.

(B) The balanced chemical equation is:
$Ba(OH)_2 + CO_2 \rightarrow BaCO_3 + H_2O$
From the stoichiometry,$1 \ mol$ of $Ba(OH)_2$ produces $1 \ mol$ of $BaCO_3$.
Therefore,$0.205 \ mol$ of $Ba(OH)_2$ will produce $0.205 \ mol$ of $BaCO_3$.
The molar mass of $BaCO_3 = 137 + 12 + (3 \times 16) = 197 \ g \ mol^{-1}$.
Mass of $BaCO_3 = \text{moles} \times \text{molar mass} = 0.205 \ mol \times 197 \ g \ mol^{-1} = 40.385 \ g \approx 40.5 \ g$ (rounding to the nearest option).

(A) The balanced chemical equation is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$
From the stoichiometry,$2 \times 23 \ g$ $(46 \ g)$ of $Na$ produces $1 \ mol$ of $H_2$ gas.
Therefore,$23 \ g$ of $Na$ will produce $\frac{1 \times 23}{46} = 0.5 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any ideal gas occupies $22400 \ mL$.
Thus,the volume of $H_2$ evolved is $0.5 \times 22400 \ mL = 11200 \ mL$.

(A) According to Gay-Lussac's Law of Gaseous Volumes,the reaction is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$.
From the stoichiometry,$1 \ volume$ of $H_2$ reacts with $1 \ volume$ of $Cl_2$ to produce $2 \ volumes$ of $HCl$.
Therefore,$40 \ mL$ of $H_2$ will react with $40 \ mL$ of $Cl_2$ to produce $2 \times 40 \ mL = 80 \ mL$ of $HCl$.
Thus,the volume of chlorine required is $40 \ mL$ and the volume of $HCl$ formed is $80 \ mL$.

(D) The combustion reaction for butane $(C_4H_{10})$ is:
$C_4H_{10} + \frac{13}{2} O_2 \to 4CO_2 + 5H_2O$
The molar mass of butane is $58 \ g/mol$.
For $1 \ kg$ $(1000 \ g)$ of butane,the number of moles is $n = \frac{1000}{58} \ mol$.
According to the stoichiometry,$1 \ mol$ of $C_4H_{10}$ requires $\frac{13}{2} \ mol$ of $O_2$.
So,moles of $O_2$ required = $\frac{13}{2} \times \frac{1000}{58} \ mol$.
Mass of $O_2$ required = $\frac{13}{2} \times \frac{1000}{58} \times 32 \ g$.
Mass of $O_2$ = $3586.2 \ g \approx 3.58 \ kg$.

(C) The balanced chemical equation for the formation of ozone is: $3O_2 \rightarrow 2O_3$.
At $STP$,$22.4 \ L$ corresponds to $1 \ mol$ of gas.
Number of moles of $O_3$ produced $= \frac{33.6 \ L}{22.4 \ L/mol} = 1.5 \ mol$.
According to the stoichiometry,$2 \ mol$ of $O_3$ are produced from $3 \ mol$ of $O_2$.
Therefore,$1.5 \ mol$ of $O_3$ requires: $\frac{3}{2} \times 1.5 = 2.25 \ mol$ of $O_2$ (theoretical).
Since the conversion efficiency is $15\%$,the actual amount of $O_2$ required is: $\frac{2.25 \ mol}{0.15} = 15 \ mol$ of $O_2$.
The molar mass of $O_2$ is $32 \ g/mol$.
Mass of $O_2$ required $= 15 \ mol \times 32 \ g/mol = 480 \ g$.

(B) The chemical reaction is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$
The molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
The molar mass of $CaO = 40 + 16 = 56 \ g/mol$.
According to the stoichiometry of the reaction,$100 \ g$ of $CaCO_3$ produces $56 \ g$ of $CaO$.
Therefore,to obtain $56 \ kg$ of $CaO$,we require $100 \ kg$ of $CaCO_3$.

(A) The chemical equation for the dehydration of cyclohexanol is: $C_6H_{11}OH \rightarrow C_6H_{10} + H_2O$
$1$. Molar mass of cyclohexanol $(C_6H_{11}OH) = (6 \times 12) + (12 \times 1) + 16 = 72 + 12 + 16 = 100 \ g/mol$.
$2$. Molar mass of cyclohexene $(C_6H_{10}) = (6 \times 12) + (10 \times 1) = 72 + 10 = 82 \ g/mol$.
$3$. Moles of cyclohexanol taken $= \frac{100 \ g}{100 \ g/mol} = 1 \ mol$.
$4$. Theoretical yield of cyclohexene $= 1 \ mol \times 82 \ g/mol = 82 \ g$.
$5$. Actual yield with $75\%$ efficiency $= 82 \ g \times 0.75 = 61.5 \ g$.

(D) The balanced chemical equation for the reaction is:
$CO_2(g) + C(s) \rightarrow 2CO(g)$
According to the stoichiometry of the reaction,$1 \ volume$ of $CO_2$ produces $2 \ volumes$ of $CO$.
Therefore,$26 \ mL$ of $CO_2$ will produce $2 \times 26 \ mL = 52 \ mL$ of $CO$.

(D) $Step \ I$: Calculate the number of oxygen atoms in $22 \ g$ of $CO_2$.
Molar mass of $CO_2 = 12 + 2 \times 16 = 44 \ g/mol$.
Moles of $CO_2 = \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of $CO_2$ contains $2 \ mol$ of oxygen atoms,$0.5 \ mol$ of $CO_2$ contains $0.5 \times 2 = 1 \ mol$ of oxygen atoms.
$Step \ II$: Calculate the weight of $CO$ containing $1 \ mol$ of oxygen atoms.
Molar mass of $CO = 12 + 16 = 28 \ g/mol$.
$1 \ mol$ of $CO$ contains $1 \ mol$ of oxygen atoms.
Therefore,the weight of $CO$ required is $28 \ g$.

(B) The formula for normality $(N)$ is given by: $N = \frac{10 \times \text{specific gravity} \times \text{percentage by weight}}{\text{equivalent weight}}$.
The equivalent weight of $H_2SO_4$ is $\frac{98}{2} = 49$.
Substituting the values: $N = \frac{10 \times 1.71 \times 80}{49}$.
$N = \frac{1368}{49} \approx 27.92 \approx 27.9 \, N$.

(A) The balanced chemical equation is: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$.
Initial moles: $SO_2 = 10 \ mol$,$O_2 = 15 \ mol$,$SO_3 = 0 \ mol$.
According to the stoichiometry,$2 \ mol$ of $SO_2$ produces $2 \ mol$ of $SO_3$.
Since $8 \ mol$ of $SO_3$ is formed,the amount of $SO_2$ consumed is $8 \ mol$.
Remaining $SO_2 = 10 - 8 = 2 \ mol$.
For $2 \ mol$ of $SO_2$,$1 \ mol$ of $O_2$ is required.
So,for $8 \ mol$ of $SO_2$,the amount of $O_2$ consumed is $8 / 2 = 4 \ mol$.
Remaining $O_2 = 15 - 4 = 11 \ mol$.
Thus,the unreacted moles are $2 \ mol$ of $SO_2$ and $11 \ mol$ of $O_2$.

(D) The balanced chemical equation is $2Al(s) + \frac{3}{2}O_2(g) \to Al_2O_3(s)$.
According to the stoichiometry of the reaction:
$2$ moles of $Al$ react with $1.5$ $(3/2)$ moles of $O_2$ to produce $1$ mole of $Al_2O_3$.

(NONE) The chemical reaction for the formation of barium bicarbonate is:
$BaCO_3 + CO_2 + H_2O \rightarrow Ba(HCO_3)_2$
According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ produces $1 \ mol$ of $Ba(HCO_3)_2$.
Therefore,$0.205 \ mol$ of $BaCO_3$ will produce $0.205 \ mol$ of $Ba(HCO_3)_2$.
The molar mass of $Ba(HCO_3)_2$ is $137 + 2 \times (1 + 12 + 3 \times 16) = 137 + 2 \times 61 = 137 + 122 = 259 \ g/mol$.
Mass $= \text{moles} \times \text{molar mass} = 0.205 \ mol \times 259 \ g/mol = 53.095 \ g$.

(D) The balanced chemical equation for the reaction is:
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
From the equation,$2 \ mol$ of $NaOH$ $(2 \times 40 = 80 \ g)$ reacts with $1 \ mol$ of $H_2SO_4$ $(98 \ g)$.
Moles of $NaOH$ in $5 \ g = \frac{5}{40} = 0.125 \ mol$.
According to the stoichiometry,$2 \ mol$ of $NaOH$ requires $1 \ mol$ of $H_2SO_4$.
Therefore,$0.125 \ mol$ of $NaOH$ requires $\frac{1}{2} \times 0.125 = 0.0625 \ mol$ of $H_2SO_4$.
Mass of pure $H_2SO_4 = 0.0625 \ mol \times 98 \ g/mol = 6.125 \ g$.
Since the $H_2SO_4$ is $90\%$ pure,the required weight is $\frac{6.125}{0.90} \approx 6.81 \ g$.

(B) The balanced chemical equation for the decomposition of potassium chlorate is:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
At $NTP$,$1 \ mole$ of any gas occupies $22.4 \ L$.
Therefore,the number of moles of $O_2$ produced is:
$n(O_2) = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
From the stoichiometry of the reaction,$3 \ moles$ of $O_2$ are produced from $2 \ moles$ of $KClO_3$.
So,$1 \ mole$ of $O_2$ is produced from $\frac{2}{3} \ moles$ of $KClO_3$.
Thus,$0.5 \ moles$ of $O_2$ will be produced from:
$n(KClO_3) = \frac{2}{3} \times 0.5 = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \ mol$.

(A) From the balanced chemical equation,$1 \ mol$ of $C_6H_{12}O_6$ is produced from $6 \ mol$ of $CO_2$.
Molar mass of $C_6H_{12}O_6 = 180 \ g \ mol^{-1}$.
Moles of $1.8 \ g$ glucose $= \frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.
Since $1 \ mol$ glucose requires $6 \ mol$ $CO_2$,then $0.01 \ mol$ glucose requires $0.01 \times 6 = 0.06 \ mol$ of $CO_2$.
Number of molecules of $CO_2 = \text{moles} \times N_A = 0.06 \times 6.022 \times 10^{23}$.

(C) Volume of the block = $10 \, cm \times 15 \, cm \times 20 \, cm = 3000 \, cm^3$.
Mass of the block = $\text{Volume} \times \text{Density} = 3000 \, cm^3 \times 8.17 \, g \, cm^{-3} = 24510 \, g$.
Mass of iron in the block = $\frac{54.7}{100} \times 24510 \, g = 13406.97 \, g$.
Number of moles of $Fe = \frac{13406.97 \, g}{55.85 \, g \, mol^{-1}} \approx 240.05 \, mol$.
Number of $Fe$ atoms = $\text{moles} \times N_A = 240.05 \times 6.022 \times 10^{23} \approx 1.445 \times 10^{26}$.
Given the options provided,the closest calculated value is $14.41 \times 10^{25}$.

(B) The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Number of moles of $CO_2 = \frac{\text{given mass}}{\text{molar mass}} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,the volume of $0.1 \ mol$ of $CO_2 = 0.1 \ mol \times 22.4 \ L/mol = 2.24 \ L$.

(C) The given reaction is $N_2 + 3H_2 \rightarrow 2NH_3$.
Given volume of $N_2 = 40 \, mL$ at $NTP$.
First,calculate the mass of $N_2$ used:
$Moles \, of \, N_2 = \frac{40 \, mL}{22400 \, mL/mol} = \frac{1}{560} \, mol$.
$Mass \, of \, N_2 = Moles \times Molar \, mass = \frac{1}{560} \times 28 \, g/mol = 0.05 \, g$.
According to the stoichiometry of the reaction,$28 \, g$ of $N_2$ produces $34 \, g$ of $NH_3$.
Therefore,the mass of $NH_3$ produced by $0.05 \, g$ of $N_2$ is:
$Mass \, of \, NH_3 = \frac{34 \, g \, NH_3}{28 \, g \, N_2} \times 0.05 \, g \, N_2 = 0.0607 \, g$.

(C) The molar mass of the mixture is calculated as $M_{mix} = 2 \times \text{Vapour Density} = 2 \times 38.3 = 76.6 \, g/mol$.
Let the number of moles of $NO_2$ be $x$. Then the number of moles of $N_2O_4$ is $(100 - x)$.
The molar mass of $NO_2$ is $46 \, g/mol$ and $N_2O_4$ is $92 \, g/mol$.
The total mass of the mixture is given by the sum of the masses of the components:
$x(46) + (100 - x)(92) = 100 \times 76.6$.
$46x + 9200 - 92x = 7660$.
$-46x = 7660 - 9200$.
$-46x = -1540$.
$x = \frac{1540}{46} \approx 33.48 \, moles$.

(A) The decomposition reaction is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$.
From the stoichiometry,$100 \, g$ of $CaCO_3$ produces $56 \, g$ of $CaO$.
Therefore,$0.56 \, g$ of $CaO$ corresponds to $\frac{100}{56} \times 0.56 = 1 \, g$ of $CaCO_3$.
The reaction between $CaCl_2$ and $Na_2CO_3$ is: $CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaCl$.
Since $111 \, g$ of $CaCl_2$ produces $100 \, g$ of $CaCO_3$,$1 \, g$ of $CaCO_3$ is produced by $\frac{111}{100} \times 1 = 1.11 \, g$ of $CaCl_2$.
The mass of $NaCl$ in the mixture is $4.44 \, g - 1.11 \, g = 3.33 \, g$.
The percentage of $NaCl$ is $\frac{3.33}{4.44} \times 100 = 75 \%$.

(C) The chemical equation is: $C_7H_{14}(\ell) \rightarrow C_7H_8(\ell) + 3H_2(g)$.
Calculate the molar masses:
$M(C_7H_{14}) = (7 \times 12) + (14 \times 1) = 84 + 14 = 98 \ g/mol$.
$M(C_7H_8) = (7 \times 12) + (8 \times 1) = 84 + 8 = 92 \ g/mol$.
Since $C_7H_{14}$ and $C_7H_8$ are liquids,the initial weight of the liquid is $98 \ g$ and the final weight of the liquid is $92 \ g$.
The decrease in the weight of the liquid is $98 \ g - 92 \ g = 6 \ g$.
The percentage decrease in the weight of the liquid is calculated as:
$\text{Percentage decrease} = \frac{\text{Decrease in weight}}{\text{Initial weight}} \times 100$
$\text{Percentage decrease} = \frac{6}{98} \times 100 \approx 6.12\%$.

(B) The balanced chemical equation for the neutralization reaction is:
$2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$
Using the stoichiometry formula $\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$:
Where $M_1 = 1 \, M$,$V_1 = ?$,$n_1 = 1$ (for $H_2SO_4$)
And $M_2 = 1 \, M$,$V_2 = 10 \, ml$,$n_2 = 2$ (for $NaOH$)
$\frac{1 \times V_1}{1} = \frac{1 \times 10}{2}$
$V_1 = 5 \, ml$.

(A) The total milliequivalents of $HCl$ are: $N \times V = 0.1 \times 100 = 10 \, meq$.
The milliequivalents of $NaOH$ added are: $N \times V = 0.2 \times 30 = 6 \, meq$.
For complete neutralization,the total milliequivalents of base must equal the milliequivalents of acid:
$N_{HCl} \times V_{HCl} = (N_{NaOH} \times V_{NaOH}) + (N_{KOH} \times V_{KOH})$
Substituting the values:
$10 = 6 + (0.25 \times V_{KOH})$
$4 = 0.25 \times V_{KOH}$
$V_{KOH} = \frac{4}{0.25} = 16 \, mL$.

(D) The chemical reaction is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$.
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \, g/mol$.
Moles of $CaCO_3 = \frac{1 \, g}{100 \, g/mol} = 0.01 \, mol$.
According to the stoichiometry,$1 \, mol$ of $CaCO_3$ reacts with $2 \, mol$ of $HCl$.
Therefore,$0.01 \, mol$ of $CaCO_3$ requires $0.02 \, mol$ of $HCl$.
Using the formula $M = \frac{n}{V(L)}$,we have $V(L) = \frac{n}{M} = \frac{0.02 \, mol}{0.1 \, M} = 0.2 \, L$.
Converting to $cm^3$ (or $mL$): $0.2 \, L \times 1000 \, cm^3/L = 200 \, cm^3$.

(B) In an acidic medium,the reaction for $KMnO_4$ is:
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.
The $n$-factor for $KMnO_4$ in an acidic medium is $5$.
Equivalent weight = $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{158}{5} = 31.6 \ g/eq$.
Normality $(N)$ = $\frac{\text{Mass}}{\text{Equivalent weight} \times \text{Volume in } L}$.
$1 \ N = \frac{\text{Mass}}{31.6 \times 1 \ L}$.
Mass = $31.6 \ g$.

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

(C) The neutralization reaction is: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$.
Using the molarity equation for neutralization: $M_1V_1n_1 = M_2V_2n_2$,where $n$ is the acidity/basicity factor.
For $NaOH$,$n_1 = 1$ and for $H_2SO_4$,$n_2 = 2$.
Given: $M_1 = 0.2 \, M$,$V_1 = 40 \, mL$,$M_2 = 0.1 \, M$.
Substituting the values: $0.2 \times 40 \times 1 = 0.1 \times V_2 \times 2$.
$8 = 0.2 \times V_2$.
$V_2 = \frac{8}{0.2} = 40 \, mL$.

(C) According to the law of equivalence,the number of gram equivalents of $KMnO_4$ must be equal to the number of gram equivalents of oxalic acid.
$N_1 V_1 = N_2 V_2$
Given: $N_1 = 1 \, N$,$V_1 = 20 \, mL$,$V_2 = 20 \, mL$.
$1 \times 20 = N_2 \times 20 \implies N_2 = 1 \, N$.
Normality $(N)$ is defined as $\frac{\text{Number of equivalents}}{\text{Volume in Liters}}$.
Number of equivalents = $N \times V(L) = 1 \times 0.020 = 0.02 \, eq$.
For oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$,the n-factor is $2$.
Equivalent weight = $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{126}{2} = 63 \, g/eq$.
Weight = $\text{Number of equivalents} \times \text{Equivalent weight} = 0.02 \times 63 = 1.26 \, g$.
Wait,re-evaluating the question context: If the question implies the weight of oxalic acid in $1 \, L$ of $1 \, N$ solution,then Weight = $N \times \text{Equivalent weight} \times V(L) = 1 \times 63 \times 1 = 63 \, g$.

(A) The theoretical weight $(W_{th})$ is calculated based on the assumption that the substance is $100\%$ pure.
Since the sample is impure,its purity is less than $100\%$.
To obtain the required amount of the pure substance,we must take a larger quantity of the impure sample.
Therefore,the actual weight required $(W_{act})$ is calculated as $W_{act} = \frac{W_{th} \times 100}{\text{Percentage Purity}}$.
Since the percentage purity is less than $100$,$W_{act}$ will be greater than $W_{th}$.

(C) For neutralization,the number of equivalents of acid must equal the number of equivalents of base.
Equivalents of $NaOH = N \times V(L) = \frac{1}{10} \times \frac{25}{1000} = 0.0025 \, eq$.
Let the molecular weight of the dibasic acid be $M$.
The equivalent weight of the dibasic acid is $E = \frac{M}{2}$ (since it is dibasic).
Equivalents of acid = $\frac{\text{mass}}{E} = \frac{0.16}{M/2} = \frac{0.32}{M}$.
Equating the two: $\frac{0.32}{M} = 0.0025$.
$M = \frac{0.32}{0.0025} = 128$.

(A) The molar mass of $SO_2 = 32 + (2 \times 16) = 64 \ g/mol$.
At $STP$,the molar volume of an ideal gas is $22.4 \ L/mol$.
Density $(\rho) = \frac{\text{Molar Mass}}{\text{Molar Volume}}$.
$\rho = \frac{64 \ g/mol}{22.4 \ L/mol} \approx 2.86 \ g/L$.

(C) The decomposition reaction is: $CaCO_3(s) \to CaO(s) + CO_2(g)$.
From the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.
Using the ideal gas law $PV = nRT$,we calculate the moles of $CO_2$:
$n = \frac{PV}{RT} = \frac{(750/760) \ atm \times 20 \ L}{0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 300 \ K} \approx 0.803 \ mol$.
Rounding to the nearest significant value,$n \approx 0.8 \ mol$.
Mass of $CaCO_3 = n \times \text{Molar Mass} = 0.8 \ mol \times 100 \ g/mol = 80 \ g$.

(A) Moles of first gas $n_1 = \frac{w_1}{m_1} = \frac{0.45}{60} = 0.0075 \ mol$.
Moles of second gas $n_2 = \frac{w_2}{m_2} = \frac{0.22}{44} = 0.0050 \ mol$.
Total moles $n_{total} = n_1 + n_2 = 0.0075 + 0.0050 = 0.0125 \ mol$.
Partial pressure of the second gas $P_2 = \left( \frac{n_2}{n_{total}} \right) \times P_{total} = \left( \frac{0.0050}{0.0125} \right) \times 75 = 30 \ cm \ of \ Hg$.

(D) Let the mass of both $CH_4$ and $H_2$ be $m \, g$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $m / 16$.
Number of moles of $H_2$ $(n_{H_2})$ = $m / 2$.
Total moles $(n_{total})$ = $n_{CH_4} + n_{H_2} = m/16 + m/2 = (m + 8m) / 16 = 9m / 16$.
The partial pressure fraction of a gas is equal to its mole fraction.
Mole fraction of $H_2$ $(x_{H_2})$ = $n_{H_2} / n_{total} = (m/2) / (9m/16) = (m/2) \times (16/9m) = 8/9$.
Therefore,the fraction of total pressure exerted by hydrogen is $8/9$.

(D) The balanced chemical equation for the reduction of boron trichloride by hydrogen is:
$2BCl_3(g) + 3H_2(g) \rightarrow 2B(s) + 6HCl(g)$
From the stoichiometry,$2 \ mol$ of $B$ are produced by $3 \ mol$ of $H_2$.
Moles of $B$ produced = $\frac{\text{mass}}{\text{atomic mass}} = \frac{21.6 \ g}{10.8 \ g/mol} = 2 \ mol$.
Since $2 \ mol$ of $B$ require $3 \ mol$ of $H_2$,the moles of $H_2$ required is $3 \ mol$.
At $STP$ ($273 \ K$ and $1 \ atm$),$1 \ mol$ of any ideal gas occupies $22.4 \ L$.
Therefore,volume of $H_2$ = $3 \ mol \times 22.4 \ L/mol = 67.2 \ L$.

(C) Let the mass of both $CH_4$ and $O_2$ be $w \, g$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{w}{16}$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{w}{32}$.
According to Dalton's Law of partial pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure.
Mole fraction of $O_2$ $(x_{O_2})$ = $\frac{n_{O_2}}{n_{CH_4} + n_{O_2}} = \frac{w/32}{w/16 + w/32}$.
$x_{O_2} = \frac{w/32}{2w/32 + w/32} = \frac{w/32}{3w/32} = \frac{1}{3}$.
Therefore,the fraction of total pressure exerted by oxygen is $1/3$.

(C) According to Avogadro's Law,at constant temperature and pressure,the number of moles $(n)$ of gas is proportional to the volume $(V)$ of the container. Since the volume of the flask is constant,the number of moles of $CH_4$ and $O_2$ will be equal.
$n_{CH_4} = n_{O_2}$
$\frac{w_{CH_4}}{M_{CH_4}} = \frac{w_{O_2}}{M_{O_2}}$
Given $M_{CH_4} = 16 \ g/mol$ and $M_{O_2} = 32 \ g/mol$,we have:
$\frac{w_{CH_4}}{16} = \frac{w_{O_2}}{32}$
$w_{O_2} = 2 \times w_{CH_4}$
Thus,the mass of oxygen is double the mass of methane.

(D) The molar mass of $HI$ is $1 \times 1 + 127 = 128 \ g/mol$.
Number of moles of $HI = \frac{\text{mass}}{\text{molar mass}} = \frac{64 \ g}{128 \ g/mol} = 0.5 \ mol$.
Active mass (molar concentration) = $\frac{\text{moles}}{\text{volume in liters}} = \frac{0.5 \ mol}{2 \ L} = 0.25 \ mol/L$.

(D) Active mass is defined as molar concentration (molarity) in $mol/L$.
Formula: $\text{Active Mass} = \frac{\text{Mass in grams}}{\text{Molar mass} \times \text{Volume in Liters}}$.
Given density $d = \frac{m}{V}$,so $m = d \times V$.
For $1 \, L$ $(1000 \, mL)$ of solution:
$1$. For $CH_3OH$ $(d = 0.5 \, g/mL)$:
Mass $= 0.5 \, g/mL \times 1000 \, mL = 500 \, g$.
$Molar mass of CH_3OH = 12 + 3(1) + 16 + 1 = 32 \, g/mol$.
Active mass $= \frac{500 \, g}{32 \, g/mol \times 1 \, L} = 15.625 \approx 15.62 \, mol/L$.
$2$. For $CCl_4$ $(d = 1.3 \, g/mL)$:
Mass $= 1.3 \, g/mL \times 1000 \, mL = 1300 \, g$.
$Molar mass of CCl_4 = 12 + 4(35.5) = 12 + 142 = 154 \, g/mol$.
Active mass $= \frac{1300 \, g}{154 \, g/mol \times 1 \, L} = 8.44 \, mol/L$.
Note: The provided option $D$ ($15.62$ and $7.79$) assumes a density of $1.2 \, g/mL$ for $CCl_4$ $(1200/154 = 7.79)$. Based on the provided density of $1.3 \, g/mL$,the calculation yields $8.44$. Given the options,$D$ is the intended answer based on the calculation method.