Chemical stoichiometry Questions in English

(A) The theoretical weight is calculated based on $100\%$ purity of the substance.
Since the sample is impure,its percentage purity is less than $100\%$.
To obtain the required amount of the pure compound,we must take a larger quantity of the impure sample to compensate for the impurities present.
Therefore,the weight of the impure substance required will be greater than the theoretical weight calculated for a pure substance.

(C) At $STP$,$1 \ mole$ of any ideal gas occupies $22.4 \ L$ of volume.
Given that $5.8 \ L$ of the gas has a mass of $7.5 \ g$.
Therefore,the mass of $22.4 \ L$ (molar mass) of the gas is calculated as:
$\text{Molar mass} = \frac{7.5 \ g}{5.8 \ L} \times 22.4 \ L/mol \approx 28.96 \ g/mol \approx 29 \ g/mol$.
The molar masses of the given options are:
$NO = 14 + 16 = 30 \ g/mol$
$N_2O = 28 + 16 = 44 \ g/mol$
$CO = 12 + 16 = 28 \ g/mol$
$CO_2 = 12 + 32 = 44 \ g/mol$
The calculated molar mass of $29 \ g/mol$ is closest to $CO$ $(28 \ g/mol)$.

(A) At $STP$,the volume of $1 \ mol$ of any ideal gas is $22.4 \ L$.
Given that $1 \ L$ of the gas weighs $1.16 \ g$.
Therefore,the molar mass of the gas is the weight of $22.4 \ L$ of the gas at $STP$.
Molar mass $= 1.16 \ g/L \times 22.4 \ L/mol = 25.984 \ g/mol \approx 26 \ g/mol$.
Comparing this with the molar masses of the options:
$C_2H_2 = (2 \times 12) + (2 \times 1) = 26 \ g/mol$.
$CO = 12 + 16 = 28 \ g/mol$.
$O_2 = 2 \times 16 = 32 \ g/mol$.
$CH_4 = 12 + (4 \times 1) = 16 \ g/mol$.
Thus,the gas is $C_2H_2$.

(D) The reaction between a dibasic acid $(H_2A)$ and barium hydroxide $(Ba(OH)_2)$ is: $H_2A + Ba(OH)_2 \rightarrow BaA + 2H_2O$.
From the stoichiometry,$1 \text{ mole}$ of $Ba(OH)_2$ reacts with $1 \text{ mole}$ of dibasic acid.
Number of moles of $Ba(OH)_2 = \text{Molarity} \times \text{Volume (in L)} = 0.25 \ M \times 0.025 \ L = 0.00625 \ mol$.
Since the acid is dibasic,the number of moles of acid is also $0.00625 \ mol$.
Molecular mass of acid $= \frac{\text{Mass}}{\text{Moles}} = \frac{1.25 \ g}{0.00625 \ mol} = 200 \ g/mol$.

(D) The balanced chemical equation for the decomposition of $BaCO_3$ is:
$BaCO_3(s) \to BaO(s) + CO_2(g)$
The molar mass of $BaCO_3 = 137 + 12 + (3 \times 16) = 197 \ g/mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ $(197 \ g)$ produces $1 \ mol$ of $CO_2$ gas,which occupies $22.4 \ L$ at $STP$.
Therefore,the volume of $CO_2$ produced from $9.85 \ g$ of $BaCO_3$ is:
$V = \frac{22.4 \ L}{197 \ g} \times 9.85 \ g = 1.12 \ L$.

(B) The balanced chemical equation for the combustion of ethylene $(C_2H_4)$ is:
$C_2H_4 + 3O_2 \to 2CO_2 + 2H_2O$
From the stoichiometry of the reaction:
$1 \, \text{mole}$ of $C_2H_4$ $(28 \, g)$ requires $3 \, \text{moles}$ of $O_2$ $(3 \times 32 = 96 \, g)$.
Therefore,$28 \, g$ of $C_2H_4$ requires $96 \, g$ of $O_2$.
For $2.8 \, kg$ $(2800 \, g)$ of $C_2H_4$,the mass of $O_2$ required is:
$\text{Mass of } O_2 = \frac{96 \, g}{28 \, g} \times 2800 \, g = 9600 \, g = 9.6 \, kg$.

(A) The balanced chemical equation for the decomposition of phosphine is:
$2PH_{3(g)} \to 2P_{(s)} + 3H_{2(g)}$
According to the stoichiometry of the reaction,$2 \ mL$ of $PH_3$ gas produces $3 \ mL$ of $H_2$ gas,while phosphorus is formed as a solid.
For $100 \ mL$ of $PH_3$,the volume of $H_2$ produced is:
$V_{H_2} = \frac{3}{2} \times 100 \ mL = 150 \ mL$
The change in volume is the difference between the final volume of gas and the initial volume of gas:
$\text{Increase in volume} = 150 \ mL - 100 \ mL = 50 \ mL$.

(B) The balanced chemical equation for the reaction is: $Mg + 2HCl \to MgCl_2 + H_2$
From the stoichiometry of the reaction,$1 \ mol$ of $Mg$ $(24 \ g)$ produces $1 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ volume.
Therefore,$24 \ g$ of $Mg$ produces $22.4 \ L$ of $H_2$ at $STP$.
For $12 \ g$ of $Mg$,the volume of $H_2$ produced is: $\frac{22.4 \ L}{24 \ g} \times 12 \ g = 11.2 \ L$.

(A) At standard temperature and pressure $(STP)$,$22.4 \ L$ of any gas corresponds to $1 \ mole$ of the gas.
Given that $2.24 \ L$ of the gas has a mass of $4.4 \ g$.
Therefore,the mass of $22.4 \ L$ (which is $1 \ mole$) of the gas is calculated as:
$\text{Molar mass} = \frac{4.4 \ g}{2.24 \ L} \times 22.4 \ L/mol = 44 \ g/mol$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Thus,the gas is $CO_2$.

(D) The density of a gas at $STP$ is calculated using the formula: $\text{Density} = \frac{\text{Molecular weight}}{\text{Molar volume at STP}}$.
Given,molecular weight $= 45$ and molar volume at $STP = 22.4 \ L \ mol^{-1}$.
Therefore,$\text{Density} = \frac{45}{22.4} \approx 2.01 \ g \ L^{-1}$.
Thus,the correct option is $D$.

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

(C) The chemical reaction between calcium phosphide $(Ca_3P_2)$ and water is as follows:
$Ca_3P_2 + 6H_2O \to 2PH_3 + 3Ca(OH)_2$
According to the stoichiometry of the balanced chemical equation,$1 \text{ mole}$ of $Ca_3P_2$ reacts with $6 \text{ moles}$ of $H_2O$ to produce $2 \text{ moles}$ of phosphine $(PH_3)$ and $3 \text{ moles}$ of calcium hydroxide $(Ca(OH)_2)$.
Therefore,the correct option is $C$.

(C) The density of liquid $HCl$ is $1.17 \ g/cc$.
This means $1 \ cc$ of $HCl$ has a mass of $1.17 \ g$.
Therefore,$1000 \ cc$ $(1 \ L)$ of $HCl$ has a mass of $1170 \ g$.
The molar mass of $HCl$ is $1.008 + 35.45 = 36.458 \ g/mol$,which is approximately $36.5 \ g/mol$.
Molarity = $\frac{\text{mass of solute in grams}}{\text{molar mass} \times \text{volume of solution in litres}}$.
Molarity = $\frac{1170 \ g}{36.5 \ g/mol \times 1 \ L} = 32.05 \ mol/L$.

(B) The neutralization reaction is $NaOH + HCl \rightarrow NaCl + H_2O$.
For neutralization,the number of equivalents of $NaOH$ must equal the number of equivalents of $HCl$.
Number of equivalents of $HCl = N \times V \text{ (in } L) = 0.1 \ N \times 1.5 \ L = 0.15 \ \text{equivalents}$.
Since the $n$-factor of $NaOH$ is $1$,its equivalent weight is equal to its molar mass,which is $23 + 16 + 1 = 40 \ g/mol$.
Mass of $NaOH = \text{Number of equivalents} \times \text{Equivalent weight} = 0.15 \times 40 = 6 \ g$.

(A) The chemical equation for the thermal decomposition of silver carbonate is:
$2Ag_2CO_3 \xrightarrow{\Delta} 4Ag + 2CO_2 + O_2$
The molar mass of $Ag_2CO_3$ is $(2 \times 108) + 12 + (3 \times 16) = 276 \, g/mol$.
According to the stoichiometry of the reaction,$2 \, mol$ of $Ag_2CO_3$ $(2 \times 276 = 552 \, g)$ produces $4 \, mol$ of solid silver residue $(4 \times 108 = 432 \, g)$.
Therefore,$2.76 \, g$ of $Ag_2CO_3$ will yield:
$\text{Mass of Ag} = \frac{432 \, g \times 2.76 \, g}{552 \, g} = 2.16 \, g$.

(C) Given that $100 \ g$ of haemoglobin $(Hb)$ contains $0.33 \ g$ of iron $(Fe)$.
Therefore,the mass of $Fe$ in $67200 \ g$ of $Hb$ is calculated as:
$\text{Mass of } Fe = \frac{67200 \times 0.33}{100} = 221.76 \ g$.
The number of iron atoms in one molecule of haemoglobin is given by the ratio of the mass of $Fe$ in one molecule to the atomic weight of $Fe$:
$\text{Number of } Fe \text{ atoms} = \frac{221.76}{56} = 3.96 \approx 4$.
Thus,there are $4$ iron atoms present in one molecule of haemoglobin.

(C) The chemical reaction for the production of $NH_3$ from ammonium sulphate is: $(NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3 + 2H_2O$.
The neutralization reaction is: $NH_3 + HCl \rightarrow NH_4Cl$.
Combining these,the stoichiometry is: $(NH_4)_2SO_4 \equiv 2NH_3 \equiv 2HCl$.
Given molar masses: $(NH_4)_2SO_4 = 132 \ g/mol$ and $HCl = 36.5 \ g/mol$.
From the stoichiometry,$1 \ mol$ of $(NH_4)_2SO_4$ reacts with $2 \ mol$ of $HCl$.
$132 \ g$ of $(NH_4)_2SO_4$ reacts with $2 \times 36.5 \ g = 73 \ g$ of $HCl$.
For $292 \ g$ of $HCl$,the amount of $(NH_4)_2SO_4$ required is: $(132 / 73) \times 292 = 132 \times 4 = 528 \ g$.

(B) The balanced chemical equation for the reaction is: $2Al + \frac{3}{2}O_2 \to Al_2O_3$.
According to the stoichiometry of the reaction,$\frac{3}{2}$ moles of $O_2$ react with $2$ moles of $Al$.
Given that $1\frac{1}{2}$ moles (which is $\frac{3}{2}$ moles) of $O_2$ are used,the amount of $Al$ required is $2$ moles.
The molar mass of $Al$ is $27 \ g/mol$.
Therefore,the weight of $Al = 2 \ mol \times 27 \ g/mol = 54 \ g$.

(D) The balanced chemical equation for the reaction is:
$2Al + 2NaOH + 6H_2O \to 2Na[Al(OH)_4] + 3H_2$
From the stoichiometry of the reaction,$2 \ mol$ of $Al$ $(2 \times 27 \ g = 54 \ g)$ produces $3 \ mol$ of $H_2$.
Therefore,$27 \ g$ of $Al$ $(1 \ mol)$ will produce $\frac{3}{2} \times 1 = 1.5 \ mol$ of $H_2$.
Volume of $H_2$ at $STP = 1.5 \ mol \times 22.4 \ L/mol = 33.6 \ L$.

(D) The balanced chemical equation for the combustion of sulphur is:
$S(s) + O_2(g) \to SO_2(g)$
From the stoichiometry of the reaction,$1 \ mol$ of $O_2$ produces $1 \ mol$ of $SO_2$.
The molar mass of $SO_2$ is $32 + (2 \times 16) = 64 \ g/mol$.
Therefore,$1 \ mol$ of $O_2$ produces $64 \ g$ of $SO_2$.
For $5.0 \ mol$ of $O_2$,the weight of $SO_2$ produced is:
$5.0 \ mol \times 64 \ g/mol = 320 \ g$.

(D) The normality $(N)$ of a solution is given by the formula $N = M \times n$-factor.
For $H_3PO_4$,the basicity is $3$ (it is a tribasic acid),so its $n$-factor is $3$.
Given molarity $(M)$ = $1 \, M$.
Therefore,$N = 1 \times 3 = 3 \, N$.

(B) The relationship between normality $(N)$ and molarity $(M)$ is given by the formula: $N = M \times \text{n-factor}$.
For sulphuric acid $(H_2SO_4)$,the basicity (n-factor) is $2$ because it can provide $2$ protons ($H^+$ ions) per molecule.
Given molarity $(M)$ = $2 \ M$.
Therefore,$N = 2 \times 2 = 4 \ N$.

(A) For a dibasic acid,the equivalent weight $E$ is calculated as $E = \frac{\text{Molecular weight}}{2} = \frac{200}{2} = 100 \ g/eq$.
Decinormal strength means the normality $N = \frac{1}{10} = 0.1 \ N$.
The formula for normality is $N = \frac{W \times 1000}{E \times V(mL)}$,where $W$ is the weight in grams.
Substituting the values: $0.1 = \frac{W \times 1000}{100 \times 100}$.
$0.1 = \frac{W \times 1000}{10000} \implies 0.1 = \frac{W}{10}$.
$W = 0.1 \times 10 = 1 \ g$.

(B) The formula for normality $(N)$ given weight percentage and specific gravity is:
$N = \frac{10 \times \text{specific gravity} \times \text{weight percentage}}{\text{equivalent weight}}$
Given:
Specific gravity = $1.71$
Weight percentage = $80\%$
Equivalent weight of $H_2SO_4 = \frac{\text{Molecular weight}}{2} = \frac{98}{2} = 49$
Calculation:
$N = \frac{10 \times 1.71 \times 80}{49}$
$N = \frac{1368}{49} \approx 27.91$
Therefore,the normality is approximately $27.9$.

(D) For the neutralization reaction between $NaOH$ and $HCl$,we use the formula $N_1V_1 = N_2V_2$.
Given for $NaOH$: $N_1 = M/10 = 0.1 \, N$ and $V_1 = 20 \, mL$.
Given for $HCl$: $N_2 = M/20 = 0.05 \, N$.
Substituting the values: $0.1 \times 20 = 0.05 \times V_2$.
$2 = 0.05 \times V_2$.
$V_2 = 2 / 0.05 = 40 \, mL$.

(A) Let the volume of solution $A$ be $x \,L$. Then the volume of solution $B$ is $(2 - x) \,L$.
Using the normality equation $N_1V_1 + N_2V_2 = N_3V_3$:
$0.5x + 0.1(2 - x) = 0.2 \times 2$
$0.5x + 0.2 - 0.1x = 0.4$
$0.4x = 0.2$
$x = 0.5 \,L$ (Volume of $A$)
Volume of $B = 2 - 0.5 = 1.5 \,L$.
Thus,$0.5 \,L$ of $A$ and $1.5 \,L$ of $B$ are required.

(D) The formula for the normality of a mixture is $N_{mix}V_{mix} = N_1V_1 + N_2V_2 + N_3V_3$.
Given:
$N_1 = 1, V_1 = 5 \, mL$
$N_2 = 1/2, V_2 = 20 \, mL$
$N_3 = 1/3, V_3 = 30 \, mL$
$V_{mix} = 1000 \, mL$
Substituting the values:
$N_{mix} \times 1000 = (1 \times 5) + (1/2 \times 20) + (1/3 \times 30)$
$N_{mix} \times 1000 = 5 + 10 + 10 = 25$
$N_{mix} = 25 / 1000 = 0.025 = 1/40 = N/40$.

(B) From the balanced chemical equation: $2 \ mol$ of $MnO_4^-$ reacts with $5 \ mol$ of $C_2O_4^{2-}$.
Using the stoichiometry relation: $\frac{n_{MnO_4^-}}{2} = \frac{n_{C_2O_4^{2-}}}{5}$.
Given: $V_{MnO_4^-} = 20 \ mL$,$M_{MnO_4^-} = 0.1 \ M$.
$n_{MnO_4^-} = M \times V = 0.1 \times 20 = 2 \ mmol$.
Therefore,$n_{C_2O_4^{2-}} = \frac{5}{2} \times n_{MnO_4^-} = 2.5 \times 2 = 5 \ mmol$.
Checking option $(B)$: $50 \ mL$ of $0.1 \ M$ $H_2C_2O_4$ gives $n = 50 \times 0.1 = 5 \ mmol$. This matches the requirement.

(C) Phenolphthalein indicates the partial neutralization of $Na_2CO_3$ to $NaHCO_3$.
$Meq. \text{ of } Na_2CO_3 + Meq. \text{ of } NaOH = Meq. \text{ of } HCl$
$\frac{W}{E} \times 1000 + \frac{W}{E} \times 1000 = N \times V$
(Suppose $Na_2CO_3 = a \, g$,$NaOH = b \, g$)
$\frac{a}{53} \times 1000 + \frac{b}{40} \times 1000 = 300 \times 0.1$ ..... $(1)$
From the given data,$a = Na_2CO_3 = 0.53 \, g$.

(B) The balanced chemical equation for the reaction is:
$Ba(OH)_2 + CO_2 \to BaCO_3 + H_2O$
From the stoichiometry of the reaction,$1 \ mol$ 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 $BaCO_3 = 137 + 12 + (3 \times 16) = 197 \ g/mol$.
The mass of $BaCO_3$ produced = $\text{moles} \times \text{molar mass} = 0.205 \ mol \times 197 \ g/mol = 40.385 \ g$.
Rounding to the nearest provided option,the mass is approximately $40.5 \ g$.

(B) For $AgNO_3$: The reaction is $2AgNO_3 + H_2S \rightarrow Ag_2S + 2HNO_3$.
$2 \ mol$ of $AgNO_3$ react with $1 \ mol$ of $H_2S$.
Amount of $AgNO_3 = 100 \ mL \times 1 \ M = 100 \ mmol$.
$H_2S$ required $= \frac{100}{2} = 50 \ mmol$.
For $CuSO_4$: The reaction is $CuSO_4 + H_2S \rightarrow CuS + H_2SO_4$.
$1 \ mol$ of $CuSO_4$ reacts with $1 \ mol$ of $H_2S$.
Amount of $CuSO_4 = 100 \ mL \times 1 \ M = 100 \ mmol$.
$H_2S$ required $= 100 \ mmol$.
Ratio of $H_2S$ required $= \frac{50}{100} = \frac{1}{2}$ or $1:2$.

(C) The chemical reactions involved are:
$2CuSO_4 + 4KI \to 2CuI_2 + 2K_2SO_4$
$2CuI_2 \to Cu_2I_2 + I_2$
$I_2 + 2Na_2S_2O_3 \to 2NaI + Na_2S_4O_6$
From the stoichiometry,$1 \text{ mole of } CuSO_4 \cdot 5H_2O$ produces $0.5 \text{ mole of } I_2$,which reacts with $1 \text{ mole of } Na_2S_2O_3$ (hypo).
Thus,the equivalent weight of $CuSO_4 \cdot 5H_2O$ is equal to its molecular weight $(250 \ g/eq)$.
Equivalents of hypo = Equivalents of $CuSO_4 \cdot 5H_2O$
$N \times V(L) = \frac{x}{\text{Equivalent weight}}$
$0.1 \times \frac{100}{1000} = \frac{x}{250}$
$0.01 = \frac{x}{250}$
$x = 2.5 \ g$.

(D) The balanced chemical equation for the neutralization reaction is:
$HNO_3 + KOH \to KNO_3 + H_2O$
First,calculate the number of moles of $HNO_3$:
Molar mass of $HNO_3 = 1 + 14 + (3 \times 16) = 63 \, g/mol$
Moles of $HNO_3 = \frac{12.6 \, g}{63 \, g/mol} = 0.2 \, mol$
According to the stoichiometry of the reaction,$1 \, mol$ of $HNO_3$ reacts with $1 \, mol$ of $KOH$.
Therefore,$0.2 \, mol$ of $HNO_3$ requires $0.2 \, mol$ of $KOH$.
Now,calculate the mass of $KOH$:
Molar mass of $KOH = 39 + 16 + 1 = 56 \, g/mol$
Mass of $KOH = 0.2 \, mol \times 56 \, g/mol = 11.2 \, g$.

(A) Isobutane and $n$-butane are isomers with the same molecular formula,$C_4H_{10}$.
The balanced chemical equation for the combustion of $C_4H_{10}$ is:
$C_4H_{10} + \frac{13}{2} O_2 \to 4CO_2 + 5H_2O$
According to the stoichiometry,$1 \ mol$ of $C_4H_{10}$ $(58 \ g)$ requires $\frac{13}{2} \ mol$ of $O_2$ $(6.5 \times 32 = 208 \ g)$.
For $5 \ kg$ $(5000 \ g)$ of $C_4H_{10}$,the mass of $O_2$ required is:
$\text{Mass of } O_2 = \frac{5000 \ g \times 208 \ g}{58 \ g} \approx 17931 \ g = 17.93 \ kg$.
Thus,the required amount is approximately $17.9 \ kg$.

(B) The electrolysis of water is represented by the equation: $2H_2O(l) \to 2H_2(g) + O_2(g)$.
The total moles of gas produced is $n_{total} = \frac{16.8 \ L}{22.4 \ L/mol} = 0.75 \ mol$.
According to the stoichiometry,$2 \ mol$ of $H_2$ and $1 \ mol$ of $O_2$ are produced from $2 \ mol$ of $H_2O$. Thus,the ratio of $H_2:O_2$ is $2:1$.
Let the moles of $O_2$ be $x$. Then the moles of $H_2$ are $2x$.
$x + 2x = 0.75 \implies 3x = 0.75 \implies x = 0.25 \ mol$ of $O_2$.
Therefore,moles of $H_2 = 0.5 \ mol$.
From the balanced equation,$2 \ mol$ of $H_2$ are produced from $2 \ mol$ of $H_2O$,so $0.5 \ mol$ of $H_2$ are produced from $0.5 \ mol$ of $H_2O$.
Mass of $H_2O = 0.5 \ mol \times 18 \ g/mol = 9 \ g$.

(A) The chemical reaction for the formation of ozone is: $3O_2(g) \to 2O_3(g)$.
Given initial volume of $O_2 = 150 \ mL$.
$10\%$ of $O_2$ is converted to $O_3$,so volume of $O_2$ reacted = $150 \times 0.10 = 15 \ mL$.
According to the stoichiometry,$3 \ mL$ of $O_2$ produces $2 \ mL$ of $O_3$.
Therefore,$15 \ mL$ of $O_2$ will produce $\frac{2}{3} \times 15 = 10 \ mL$ of $O_3$.
Volume of remaining $O_2 = 150 - 15 = 135 \ mL$.
Total volume of the gaseous mixture = $135 \ mL \ (O_2) + 10 \ mL \ (O_3) = 145 \ mL$.
Turpentine oil absorbs $O_3$ gas,so the volume of the remaining gas $(O_2)$ = $135 \ mL$.

(B) The chemical reaction is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$.
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
From the stoichiometry,$1 \ mol$ of $CaCO_3$ $(100 \ g)$ reacts with $2 \ mol$ of $HCl$ ($2 \times 36.5 = 73 \ g$ of pure $HCl$).
Since we have $100 \ g$ of $CaCO_3$,we need $73 \ g$ of pure $HCl$.
The $HCl$ solution is $50\%$ pure by weight,meaning $50 \ g$ of pure $HCl$ is present in $100 \ g$ of the solution.
Therefore,the weight of the $50\%$ $HCl$ solution required = $\frac{73 \ g}{0.50} = 146 \ g$.

(B) The chemical reaction for the heating of limestone is:
$CaCO_3(s) \to CaO(s) + CO_2(g)$
Given mass of limestone = $10 \ g$.
Since it is $90\%$ pure,the mass of pure $CaCO_3 = 10 \times 0.90 = 9 \ g$.
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Number of moles of $CaCO_3 = \frac{9 \ g}{100 \ g/mol} = 0.09 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.
Therefore,$0.09 \ mol$ of $CaCO_3$ will produce $0.09 \ mol$ of $CO_2$.
At $NTP$,the volume of $1 \ mol$ of any gas is $22.4 \ L$.
Volume of $CO_2 = 0.09 \ mol \times 22.4 \ L/mol = 2.016 \ L$.

(B) The precipitation reactions are:
$Cd^{2+} + H_2S \rightarrow CdS + 2H^+$
$Cu^{2+} + H_2S \rightarrow CuS + 2H^+$
For $Cd(NO_3)_2$:
$Moles = Molarity \times Volume = 1 \ M \times 0.020 \ L = 0.020 \ mol = 20 \ mmol$.
Since the stoichiometry is $1:1$,$20 \ mmol$ of $H_2S$ is required.
For $CuSO_4$:
$Moles = Molarity \times Volume = 0.5 \ M \times 0.020 \ L = 0.010 \ mol = 10 \ mmol$.
Since the stoichiometry is $1:1$,$10 \ mmol$ of $H_2S$ is required.
The ratio of $H_2S$ required is $20:10 = 2:1$.

(B) The balanced chemical equation for the reaction of magnesium with an acid (e.g.,$HCl$) is:
$Mg + 2HCl \rightarrow MgCl_2 + H_2$
From the stoichiometry,$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$.
Number of moles of $Mg = \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{12 \ g}{24 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of $Mg$ gives $1 \ mol$ of $H_2$,$0.5 \ mol$ of $Mg$ will produce $0.5 \ mol$ (or $1/2 \ mol$) of $H_2$.

(A) The balanced chemical equation for the reaction is: $Mg + \frac{1}{2} O_2 \to MgO$.
From the stoichiometry,$0.5 \ mol$ of $O_2$ reacts with $1 \ mol$ of $Mg$.
Therefore,$1.5 \ mol$ of $O_2$ will react with $\frac{1.5}{0.5} = 3 \ mol$ of $Mg$.
The molar mass of $Mg$ is $24 \ g/mol$.
Mass of $Mg = \text{moles} \times \text{molar mass} = 3 \ mol \times 24 \ g/mol = 72 \ g$.

(A) The hardness of water is removed by reacting $Mg^{2+}$ ions with washing soda $(Na_2CO_3)$.
The reaction is: $Mg^{2+} + Na_2CO_3 \rightarrow MgCO_3 + 2Na^+$.
Number of milli-equivalents (meq) of $Mg^{2+} = \frac{\text{mass in mg}}{\text{equivalent weight}}$.
Equivalent weight of $Mg^{2+} = \frac{\text{Atomic mass}}{\text{Valency}} = \frac{24}{2} = 12$.
Number of meq of $Mg^{2+} = \frac{12.00 \ mg}{12} = 1 \text{ meq}$.
Since $1 \text{ meq}$ of $Mg^{2+}$ reacts with $1 \text{ meq}$ of $Na_2CO_3$,the required milli-equivalents of washing soda is $1$.

(C) The balanced chemical equation for the reduction of boron trichloride by hydrogen is:
$2BCl_3(g) + 3H_2(g) \to 2B(s) + 6HCl(g)$
First,calculate the number of moles of Boron $(B)$ produced:
$n(B) = \frac{\text{mass}}{\text{atomic mass}} = \frac{21.6 \ g}{10.8 \ g/mol} = 2 \ mol$
From the stoichiometry of the reaction,$2 \ mol$ of $B$ are produced by $3 \ mol$ of $H_2$ gas.
Since $2 \ mol$ of $B$ are produced,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,the volume of $H_2$ required is:
$V = 3 \ mol \times 22.4 \ L/mol = 67.2 \ L$

(D) In the metallic oxide $MO$,the oxygen atom has a valency of $-2$.
Since the formula is $MO$,the metal $M$ must have a valency of $+2$.
The phosphate ion is represented as $[PO_4]^{3-}$,which has a valency of $-3$.
To form a neutral compound,the charges must be balanced: $3 \times (+2) + 2 \times (-3) = 0$.
Therefore,the formula of the phosphate is $M_3(PO_4)_2$.

(B) The formula of the metal chloride is $MCl_2$. Since the chloride ion is $Cl^-$,the metal $M$ must have a valency of $+2$ to balance two chloride ions,i.e.,$M^{2+}$.
The phosphate ion is $PO_4^{3-}$.
To determine the formula of the phosphate of metal $M$,we use the criss-cross method:
$M^{2+}$ and $PO_4^{3-}$.
By swapping the valencies,we get $M_3(PO_4)_2$.

(A) The balanced chemical equation is: $C_{(s)} + H_2O_{(l)} \to CO_{(g)} + H_{2(g)}$
From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO$ and $1 \ mol$ of $H_2$,totaling $2 \ mol$ of gas.
Atomic mass of $C = 12 \ g/mol$.
Moles of $C = \frac{48.0 \ g}{12 \ g/mol} = 4 \ mol$.
Since $1 \ mol$ of $C$ produces $2 \ mol$ of gas,$4 \ mol$ of $C$ will produce $4 \times 2 = 8 \ mol$ of gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Total volume of gases produced = $8 \ mol \times 22.4 \ L/mol = 179.2 \ L$.

(D) At $STP$,$1 \ mole$ of any ideal gas occupies $22.4 \ L$ volume.
Given that $2.24 \ L$ of the gas has a mass of $4.4 \ g$.
Therefore,$22.4 \ L$ of the gas will have a mass of $\frac{4.4}{2.24} \times 22.4 = 44 \ g$.
The molar mass of the gas is $44 \ g/mol$.
Comparing this with the molar masses of the options:
$O_2 = 32 \ g/mol$
$CO = 28 \ g/mol$
$NO_2 = 46 \ g/mol$
$CO_2 = 44 \ g/mol$
Thus,the gas is $CO_2$.

(C) The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
At $NTP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$44 \ g$ of $CO_2$ occupies $22.4 \ L$.
Volume of $4.4 \ g$ of $CO_2 = \frac{22.4 \ L}{44 \ g} \times 4.4 \ g = 2.24 \ L$.

(B) The number of moles of $CO_2$ present in $200 \ mL$ solution is calculated as:
$n = \text{Molarity} \times \text{Volume (in L)} = 0.1 \times \frac{200}{1000} = 0.02 \ mol$.
Since $1 \ mol$ of an ideal gas occupies $22.4 \ L$ at $STP$,the volume of $0.02 \ mol$ of $CO_2$ is:
$V = 0.02 \ mol \times 22.4 \ L/mol = 0.448 \ L$.
Thus,the correct option is $(B)$.

(B) The molecular weight of the gas is calculated as: $M = 2 \times \text{Vapour Density} = 2 \times 11.2 = 22.4 \ g/mol$.
At $N.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Since the molar mass is $22.4 \ g/mol$,$22.4 \ g$ of the gas occupies $22.4 \ L$.
Therefore,the volume occupied by $11.2 \ g$ of the gas is: $\frac{22.4 \ L}{22.4 \ g} \times 11.2 \ g = 11.2 \ L$.