Chemical stoichiometry Questions in English

(B) The chemical reaction for the heating of $NaHCO_3$ is: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
$Na_2CO_3$ does not decompose at $673 \ K$.
The loss in mass is due to the release of $H_2O$ and $CO_2$.
The molar mass of $2NaHCO_3$ is $2 \times 84 = 168 \ g/mol$.
The mass of $H_2O + CO_2$ released is $18 + 44 = 62 \ g$.
Let the mass of $NaHCO_3$ be $x \ g$.
Loss in mass = $(62/168) \times x = 0.62 \ g$.
$x = (0.62 \times 168) / 62 = 1.68 \ g$.
Mass of $Na_2CO_3 = 4.0 - 1.68 = 2.32 \ g$.
Percentage of $Na_2CO_3 = (2.32 / 4.0) \times 100 = 58 \%$.

(D) The balanced chemical equation for the formation of ammonia is: $N_2 + 3H_2 \rightarrow 2NH_3$
From the stoichiometry of the reaction,$3$ moles of $H_2$ produce $2$ moles of $NH_3$.
Therefore,$1$ mole of $H_2$ produces $\frac{2}{3}$ moles of $NH_3$.
For $5$ moles of $H_2$,the amount of $NH_3$ produced is $\frac{2}{3} \times 5 = \frac{10}{3} \approx 3.33$ moles.
Thus,the correct option is $D$.

(B) The decomposition reaction of $CaCO_3$ is:
$CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$
At $\text{STP}$,heating $1 \ mol$ of $CaCO_3$ (i.e.,$100 \ g$) liberates $1 \ mol$ or $22.4 \ L$ of $CO_2$.
Mass of pure $CaCO_3$ in $10 \ g$ of $90 \%$ pure limestone $= \frac{10 \times 90}{100} = 9 \ g$.
Since $100 \ g$ of $CaCO_3$ produces $22.4 \ L$ of $CO_2$,
$9 \ g$ of $CaCO_3$ will liberate $= \frac{9 \times 22.4}{100} \ L = 2.016 \ L \approx 2.0 \ L$ of $CO_2$.

(C) The chemical formula of magnetite is $Fe_3O_4$.
The molar mass of $Fe_3O_4 = (3 \times 55.8) + (4 \times 16) = 167.4 + 64 = 231.4 \ g/mol$.
The number of moles in $600 \ g$ of $Fe_3O_4 = \frac{600}{231.4} \approx 2.593 \ mol$.
According to the stoichiometry of the reaction $Fe_3O_4 + 4CO \longrightarrow 3Fe + 4CO_2$,$1 \ mol$ of $Fe_3O_4$ produces $3 \ mol$ of $Fe$.
Therefore,$2.593 \ mol$ of $Fe_3O_4$ produces $2.593 \times 3 = 7.779 \ mol$ of $Fe$.
Mass of $Fe = 7.779 \ mol \times 55.8 \ g/mol \approx 434 \ g$.

(B) The balanced chemical equation is:
$2 H_2S_{(g)} + 3 O_{2(g)} \longrightarrow 2 SO_{2(g)} + 2 H_2O_{(g)}$
From the stoichiometry,$3 \ mol$ of $O_2$ reacts with $2 \ mol$ of $H_2S$.
At $STP$,$1 \ mol$ of gas occupies $22.4 \ L$.
Therefore,$15 \ L$ of $O_2 = \frac{15}{22.4} \ mol \approx 0.6696 \ mol$.
Using the mole ratio,the moles of $H_2S$ required = $\frac{2}{3} \times \left(\frac{15}{22.4}\right) \ mol$.
Mass of $H_2S = \text{moles} \times \text{molar mass} = \left(\frac{2}{3} \times \frac{15}{22.4}\right) \times 34 \ g \ mol^{-1}$.
Mass of $H_2S = \frac{10}{22.4} \times 34 \approx 15.178 \ g$.
The approximate mass is $15.16 \ g$.

(C) The balanced chemical equation is: $CaCO_3 + 2HCl \longrightarrow CaCl_2 + CO_2 + H_2O$
Moles of $HCl = \text{Molarity} \times \text{Volume (in L)} = 0.75 \ M \times 0.025 \ L = 0.01875 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $CaCO_3$ reacts with $2 \ mol$ of $HCl$.
Therefore,moles of $CaCO_3$ required $= \frac{0.01875}{2} = 0.009375 \ mol$.
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Mass of $CaCO_3 = 0.009375 \ mol \times 100 \ g/mol = 0.9375 \ g \approx 0.94 \ g$.

(C) The balanced chemical equation for the combustion of graphite is:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO_2$.
Since $1 \ mol$ of any substance contains $6.022 \times 10^{23}$ molecules,$1 \ mol$ of $C$ produces $6.022 \times 10^{23}$ molecules of $CO_2$.
Therefore,$0.1 \ mol$ of graphite will produce:
$0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules of $CO_2$.

(B) Let the number of moles of $Na_2CO_3$ be $x$ and $NaHCO_3$ be $x$.
The molar mass of $Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$.
The molar mass of $NaHCO_3 = 23 + 1 + 12 + (3 \times 16) = 84 \ g/mol$.
Total mass of the mixture: $106x + 84x = 5 \ g$.
$190x = 5 \implies x = \frac{5}{190} = 0.026316 \ mol$.
The reactions with $HCl$ are:
$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$ (Requires $2x$ moles of $HCl$)
$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$ (Requires $x$ moles of $HCl$)
Total moles of $HCl$ required $= 2x + x = 3x = 3 \times 0.026316 = 0.078948 \ mol$.
Volume of $0.1 \ M \ HCl$ required $= \frac{\text{moles}}{\text{molarity}} = \frac{0.078948}{0.1} = 0.78948 \ L$.
Converting to $mL$: $0.78948 \times 1000 = 789.48 \ mL \approx 789.0 \ mL$.

(D) The reaction of the metal $(M)$ with $H_2SO_4$ is: $M + H_2SO_4 \rightarrow MSO_4 + H_2$.
Total millimoles of $H_2SO_4$ taken = $50 \ mL \times 0.5 \ M = 25 \ mmol$.
Reaction with $NaOH$: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Millimoles of $NaOH$ used = $14.2 \ mL \times 1 \ M = 14.2 \ mmol$.
Since $1 \ mmol$ of $H_2SO_4$ reacts with $2 \ mmol$ of $NaOH$,the unreacted $H_2SO_4 = 14.2 / 2 = 7.1 \ mmol$.
Millimoles of $H_2SO_4$ reacted with metal = $25 - 7.1 = 17.9 \ mmol = 0.0179 \ mol$.
Since the metal has a valence of $2$,the stoichiometry is $1:1$,so moles of metal = $0.0179 \ mol$.
Atomic weight of metal = $\text{mass} / \text{moles} = 0.43 \ g / 0.0179 \ mol \approx 24.02 \ u$.

(D) According to Avogadro's Law,the volume ratio of gases is equal to the mole ratio under the same conditions of temperature and pressure.
Let the hydrocarbon be $C_xH_y$.
The combustion reaction is: $C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O$.
Given volume ratio: $V(C_xH_y) : V(CO_2) : V(H_2O) = 10 : 40 : 50 = 1 : 4 : 5$.
From the stoichiometry,$x = 4$ (moles of $CO_2$ per mole of hydrocarbon).
Also,$y/2 = 5$,which implies $y = 10$.
Therefore,the molecular formula is $C_4H_{10}$.

(D) The balanced chemical equation for the reduction of magnetite $(Fe_3O_4)$ is:
$Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$
Molar mass of $Fe_3O_4 = (3 \times 56) + (4 \times 16) = 168 + 64 = 232 \ g/mol$.
From the stoichiometry,$232 \ g$ of $Fe_3O_4$ produces $3 \times 56 = 168 \ g$ of $Fe$ at $100 \%$ efficiency.
Given the process is $80 \%$ efficient,the actual yield of $Fe$ from $232 \ g$ of $Fe_3O_4$ is $168 \times 0.8 = 134.4 \ g$.
To obtain $4 \ kg$ $(4000 \ g)$ of $Fe$,the mass of $Fe_3O_4$ required is:
$\text{Mass} = \frac{232 \times 4000}{134.4} \approx 6904.76 \ g \approx 6.9 \ kg$.

(C) The balanced chemical equation is: $C_2H_{4(g)} + H_{2(g)} \rightarrow C_2H_{6(g)}$
Calculate the molar masses:
$C_2H_4 = (2 \times 12) + (4 \times 1) = 28 \ g/mol$
$C_2H_6 = (2 \times 12) + (6 \times 1) = 30 \ g/mol$
From the stoichiometry of the reaction,$1 \ mol$ of $C_2H_4$ produces $1 \ mol$ of $C_2H_6$.
Therefore,$28 \ g$ of $C_2H_4$ produces $30 \ g$ of $C_2H_6$.
To produce $50 \ g$ of $C_2H_6$,the mass of $C_2H_4$ required is:
$\text{Mass of } C_2H_4 = \frac{28 \ g \ C_2H_4}{30 \ g \ C_2H_6} \times 50 \ g \ C_2H_6 = 46.67 \ g$.

(D) The molar mass of $Cu(NO_3)_2$ is calculated as follows:
$M = 63.5 + 2 \times (14 + 3 \times 16) = 63.5 + 2 \times (14 + 48) = 63.5 + 2 \times 62 = 63.5 + 124 = 187.5 \ g/mol$.
In $187.5 \ g$ of $Cu(NO_3)_2$,there is $63.5 \ g$ of copper.
Therefore,to obtain $1 \ g$ of copper,the amount of $Cu(NO_3)_2$ required is:
$\text{Mass} = \frac{187.5 \ g}{63.5 \ g} \times 1 \ g \approx 2.95 \ g$.
Thus,option $D$ is the correct answer.

(A) The reaction for heating $CaCO_3$ is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$.
Neutralization of $CO_2$ by $KOH$ is: $2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$.
From the stoichiometry,$2 \ mol$ of $KOH$ reacts with $1 \ mol$ of $CO_2$,which is produced from $1 \ mol$ of $CaCO_3$.
Thus,$2 \ mol$ of $KOH$ reacts with $1 \ mol$ of $CaCO_3$.
$2 \times 56 \ g$ of $KOH$ reacts with $100 \ g$ of $CaCO_3$.
$112 \ g$ of $KOH$ reacts with $100 \ g$ of $CaCO_3$.
Therefore,$28 \ g$ of $KOH$ reacts with $\frac{100 \times 28}{112} = 25 \ g$ of pure $CaCO_3$.
Percentage purity $= \frac{\text{mass of pure } CaCO_3}{\text{mass of impure } CaCO_3} \times 100 = \frac{25}{60} \times 100 = 41.66\% \approx 41.6\%$.

(D) The reaction for the heating of $NaHCO_3$ is: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
$Na_2CO_3$ does not decompose on heating.
Let the mass of $NaHCO_3$ be $x \ g$ and $Na_2CO_3$ be $y \ g$.
$x + y = 19$ (Equation $1$).
Moles of $CO_2$ produced $= \frac{1.12 \ L}{22.4 \ L/mol} = 0.05 \ mol$.
From the stoichiometry,$2 \ mol$ of $NaHCO_3$ produces $1 \ mol$ of $CO_2$.
So,moles of $NaHCO_3 = 2 \times 0.05 = 0.1 \ mol$.
Mass of $NaHCO_3 = 0.1 \ mol \times 84 \ g/mol = 8.4 \ g$.
Substituting in Equation $1$: $8.4 + y = 19 \Rightarrow y = 10.6 \ g$.

(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 $2.24 \ L \ CO_2$. Thus,$A-(iv)$.
$(B)$ $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. $106 \ g \ Na_2CO_3$ gives $22.4 \ L \ CO_2$. So,$1.06 \ g \ Na_2CO_3$ gives $0.224 \ L \ CO_2$. Thus,$B-(i)$.
$(C)$ $C + O_2 \rightarrow CO_2$. $12 \ g \ C$ gives $22.4 \ L \ CO_2$. So,$2.4 \ g \ C$ gives $(22.4 \times 2.4) / 12 = 4.48 \ L \ CO_2$. Thus,$C-(ii)$.
$(D)$ $2CO + O_2 \rightarrow 2CO_2$. $56 \ g \ CO$ gives $2 \times 22.4 \ L \ CO_2 = 44.8 \ L \ CO_2$. So,$0.56 \ g \ CO$ gives $(44.8 \times 0.56) / 56 = 0.448 \ L \ CO_2$. Thus,$D-(iii)$.
Therefore,the correct match is $A-(iv), B-(i), C-(ii), D-(iii)$.

(D) The chemical reaction for the formation of ozone is: $3O_2(g) \rightarrow 2O_3(g)$.
Let the initial volume of $O_2$ be $100 \ L$.
Let $x \ L$ of $O_2$ be converted into ozone.
According to the stoichiometry,$3 \ L$ of $O_2$ produces $2 \ L$ of $O_3$.
Therefore,$x \ L$ of $O_2$ will produce $\frac{2}{3}x \ L$ of $O_3$.
The remaining volume of $O_2$ is $(100 - x) \ L$.
The volume of $O_3$ formed is $\frac{2}{3}x \ L$.
Given that the volumes of $O_2$ and $O_3$ become equal:
$100 - x = \frac{2}{3}x$
$100 = x + \frac{2}{3}x = \frac{5}{3}x$
$x = \frac{100 \times 3}{5} = 60 \ L$.
The volume of ozone formed is $\frac{2}{3}x = \frac{2}{3} \times 60 = 40 \ L$.

(D) The thermal decomposition of calcium carbonate is given by the reaction:
$CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_2(g)$
From the stoichiometry of the reaction:
$1 \text{ mole of } CaCO_3 (100 \text{ g}) \text{ produces } 1 \text{ mole of } CaO (56 \text{ g})$.
Since $56 \text{ g}$ of $CaO$ is obtained from $100 \text{ g}$ of $CaCO_3$,
$28 \text{ g}$ of $CaO$ is obtained from:
$x = \frac{100 \times 28}{56} = 50 \text{ g}$.
Therefore,the value of $x$ is $50$.

(C) The balanced chemical equation for the combustion of sulphur is: $S(s) + O_2(g) \longrightarrow SO_2(g)$.
From the stoichiometry of the reaction,$1 \ mol$ of $S$ reacts with $1 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of any gas occupies a volume of $22.4 \ L$.
Therefore,$1 \ mol$ of $S$ requires $22.4 \ L$ of $O_2$.
For $1.5 \ mol$ of $S$,the volume of $O_2$ required is: $1.5 \ mol \times 22.4 \ L/mol = 33.6 \ L$.

(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 Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to their moles.
From the stoichiometry of the reaction,$1$ mole of $CO$ produces $1$ mole of $CO_2$.
Therefore,$1$ volume of $CO$ produces $1$ volume of $CO_2$ at $STP$.
Given that $11.207$ litres of $CO_2$ is formed,the volume of $CO$ required is also $11.207$ litres.
Thus,$X = 11.207$.

(A) The balanced chemical equation for the combustion of benzene $(C_6H_6)$ is:
$2C_6H_6(l) + 15O_2(g) \longrightarrow 12CO_2(g) + 6H_2O(l)$
Calculate the molar mass of benzene $(C_6H_6)$:
$M = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \ g/mol$
From the stoichiometry of the reaction,$2 \ mol$ of $C_6H_6$ require $15 \ mol$ of $O_2$.
Therefore,$78 \ g$ of $C_6H_6$ $(1 \ mol)$ requires $7.5 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of gas occupies $22.4 \ L$.
Volume of $O_2$ required for $78 \ g$ of $C_6H_6 = 7.5 \times 22.4 \ L = 168 \ L$.
For $39 \ g$ of $C_6H_6$ (which is $0.5 \ mol$):
Volume of $O_2 = \frac{168 \ L}{2} = 84 \ L$.

(B) The balanced chemical equation for the combustion of methane is:
$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$
From the stoichiometry,$1 \,mol$ of $CH_4$ $(16 \,g)$ requires $2 \,mol$ of $O_2$.
At $STP$,$1 \,mol$ of gas occupies $22.4 \,L$.
Therefore,$16 \,g$ of $CH_4$ requires $2 \times 22.4 \,L = 44.8 \,L$ of $O_2$.
For $32 \,g$ of $CH_4$,the volume of $O_2$ required is:
$\frac{44.8 \,L}{16 \,g} \times 32 \,g = 89.6 \,L$.

(B) The balanced chemical equation is:
$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Moles of $CaCO_3 = \frac{200 \ g}{100 \ g/mol} = 2 \ mol$.
According to the stoichiometry,$1 \ mol$ of $CaCO_3$ reacts with $2 \ mol$ of $HCl$.
Therefore,$2 \ mol$ of $CaCO_3$ requires $4 \ mol$ of $HCl$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Mass of pure $HCl$ required $= 4 \ mol \times 36.5 \ g/mol = 146 \ g$.
Given that the solution is $50\% \ (w/w)$,it means $50 \ g$ of $HCl$ is present in $100 \ g$ of solution.
Mass of solution $= \frac{146 \ g}{0.50} = 292 \ g$.

(B) The combustion reaction is: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$.
Given $4 \ g$ of $H_2$ $(n = 4/2 = 2 \ mol)$,it produces $2 \ mol$ of $H_2O$.
The density of water is $1 \ g/mL$,so the volume of $2 \ mol$ of $H_2O$ $(36 \ g)$ is $36 \ mL = 36 \times 10^{-3} \ L$.
The amount of $CO_2$ is $36 \ \mu mol = 36 \times 10^{-6} \ mol$.
Concentration of $CO_2 = \frac{n}{V} = \frac{36 \times 10^{-6} \ mol}{36 \times 10^{-3} \ L} = 10^{-3} \ M = 1 \ mM$.

(B) The volume of the spherical ball is given by $V = \frac{4}{3} \pi r^3$.
Substituting $r = 7 \ cm$,$V = \frac{4}{3} \times \frac{22}{7} \times 7^3 = \frac{4}{3} \times 22 \times 49 \approx 1437.33 \ cm^3$.
The mass of the ball is $m = V \times d = 1437.33 \ cm^3 \times 1.4 \ g \ cm^{-3} \approx 2012.26 \ g$.
The mass of iron in the ball is $56\%$ of the total mass: $m_{\text{Fe}} = 2012.26 \times 0.56 \approx 1126.87 \ g$.
The number of moles of iron is $n = \frac{m_{\text{Fe}}}{\text{atomic mass}} = \frac{1126.87}{56} \approx 20.12 \ mol$.
Thus,the number of moles of iron is approximately $20.1 \ mol$.
Therefore,option $(B)$ is correct.

(B) The formula for molarity is: $M = \frac{\text{Weight of solute (g)}}{\text{Molar mass (g/mol)}} \times \frac{1000}{\text{Volume of solution (mL)}}$
Given: $M = 0.2 \ M$,$\text{Molar mass} = 106 \ g/mol$,$\text{Volume} = 250 \ mL$.
Substituting the values: $0.2 = \frac{\text{Weight}}{106} \times \frac{1000}{250}$
$0.2 = \frac{\text{Weight}}{106} \times 4$
$\text{Weight} = \frac{0.2 \times 106}{4} = \frac{21.2}{4} = 5.3 \ g$.

(C) The molarity formula is given by $Y = \frac{X \times 1000}{m \times V}$.
Given $m = 106$,$V = 100 \ mL$,and $Y$ is the molarity.
Substituting the values: $Y = \frac{X \times 1000}{106 \times 100} = \frac{10X}{106}$.
Rearranging gives $106Y = 10X$.
For option $C$: $X = 1.06$ and $Y = 0.1$.
$106(0.1) = 10.6$ and $10(1.06) = 10.6$.
Since both sides are equal,option $C$ is correct.

(C) Step $1$: Calculate the molarity of the initial $250 \ mL$ solution.
$M = \frac{\text{mass of solute}}{\text{molar mass}} \times \frac{1000}{V(mL)}$
$M = \frac{2.65 \ g}{106 \ g/mol} \times \frac{1000}{250 \ mL} = 0.1 \ M$
Step $2$: Use the dilution formula $M_1V_1 = M_2V_2$ to find the concentration of the diluted solution.
$M_1 = 0.1 \ M, V_1 = 10 \ mL$
$V_2 = 1 \ L = 1000 \ mL$
$0.1 \times 10 = M_2 \times 1000$
$M_2 = \frac{0.1 \times 10}{1000} = 0.001 \ M$

(B) The number of molecules $(N)$ is given by the formula: $N = n \times N_A$,where $n$ is the number of moles and $N_A$ is the Avogadro constant.
Since $n = \frac{w}{M}$ (where $w$ is the mass and $M$ is the molar mass),we have $N = \frac{w}{M} \times N_A$.
Therefore,the ratio of the number of molecules of $N_2$ to $O_2$ is:
$\frac{N_{N_2}}{N_{O_2}} = \frac{w_{N_2}}{M_{N_2}} \times \frac{M_{O_2}}{w_{O_2}} = \left( \frac{w_{N_2}}{w_{O_2}} \right) \times \left( \frac{M_{O_2}}{M_{N_2}} \right)$.
Given $\frac{w_{N_2}}{w_{O_2}} = \frac{4}{1}$,$M_{N_2} = 28 \ g/mol$,and $M_{O_2} = 32 \ g/mol$.
Substituting these values: $\frac{N_{N_2}}{N_{O_2}} = \frac{4}{1} \times \frac{32}{28} = \frac{4 \times 8}{7} = \frac{32}{7}$.
Thus,the ratio is $32: 7$.

(B) To find the partial pressure,we first calculate the mole fraction of each component based on the given mass percentages.
Assume $100 \ g$ of air.
Moles of $N_2 = \frac{63 \ g}{28 \ g/mol} = 2.25 \ mol$.
Moles of $O_2 = \frac{16 \ g}{32 \ g/mol} = 0.5 \ mol$.
Moles of $Kr = \frac{21 \ g}{84 \ g/mol} = 0.25 \ mol$.
Total moles $= 2.25 + 0.5 + 0.25 = 3.0 \ mol$.
Partial pressure of a gas is given by $p_i = x_i \times p_{total}$,where $x_i$ is the mole fraction.
$p_{N_2} = (\frac{2.25}{3.0}) p = 0.75 p \ atm$.
$p_{O_2} = (\frac{0.5}{3.0}) p = 0.167 p \ atm$.
$p_{Kr} = (\frac{0.25}{3.0}) p = 0.083 p \ atm$.
Comparing these values with the given options,option $(B)$ is the closest approximation.

(D) The standard molar enthalpy of formation $(\Delta_f H^{\circ})$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$,the enthalpy change is $\Delta_r H^{\circ} = -92 \ kJ$ for the formation of $2 \ mol$ of $NH_{3(g)}$.
To find the enthalpy of formation for $1 \ mol$ of $NH_{3(g)}$,we divide the reaction by $2$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
$\Delta_f H^{\circ} = \frac{\Delta_r H^{\circ}}{2} = \frac{-92 \ kJ}{2} = -46 \ kJ \ mol^{-1}$.

(B) The charge on $1 \text{ mole}$ of electrons is $F = 96500 \text{ C/mol}$.
Given,$1 \text{ millimole} = 1 \times 10^{-3} \text{ mol}$ of $M^{n+}$ ions.
The total charge $Q = n \times \text{moles} \times F$.
$193 = n \times (1 \times 10^{-3}) \times 96500$.
$193 = n \times 96.5$.
$n = \frac{193}{96.5} = 2$.
Thus,the value of $n$ is $2$.

(C) The normality $(N)$ of the solution is given by the formula: $N = \frac{\text{Number of equivalents}}{\text{Volume in Liters}}$.
Number of equivalents $(n_{eq})$ $= N \times V(L) = 0.1 \times \frac{100}{1000} = 0.01$.
Oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ has a molecular weight of $126$. Its n-factor is $2$ (since it provides $2 \ H^+$ ions).
Equivalent mass $= \frac{\text{Molecular weight}}{n\text{-factor}} = \frac{126}{2} = 63$.
Weight required $= n_{eq} \times \text{Equivalent mass} = 0.01 \times 63 = 0.63 \ g$.

(B) The balanced chemical equation is: $HgO + 4 I^{-} + H_2O \rightarrow HgI_4^{2-} + 2 OH^{-}$.
From the stoichiometry,$1 \ mol$ of $HgO$ produces $2 \ mol$ of $OH^{-}$.
Number of moles of $HgO = \frac{0.217 \ g}{217 \ g \ mol^{-1}} = 0.001 \ mol$.
Therefore,moles of $OH^{-}$ produced $= 2 \times 0.001 = 0.002 \ mol$.
For neutralization,$n_{H^{+}} = n_{OH^{-}}$,so $n_{HCl} = 0.002 \ mol$.
Using the formula $n = M \times V(L)$,we have $0.002 = 0.01 \times V(L)$.
$V(L) = \frac{0.002}{0.01} = 0.2 \ L = 200 \ mL$.

(A) Let the mass of ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $w \ g$ each.
Molar mass of $C_2H_6 = 30 \ g/mol$.
Molar mass of $H_2 = 2 \ g/mol$.
Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{w}{30}$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{w}{2}$.
Total moles $(n_{total})$ = $n_{C_2H_6} + n_{H_2} = \frac{w}{30} + \frac{w}{2} = \frac{w + 15w}{30} = \frac{16w}{30}$.
Mole fraction of $H_2$ $(X_{H_2})$ = $\frac{n_{H_2}}{n_{total}} = \frac{w/2}{16w/30} = \frac{w}{2} \times \frac{30}{16w} = \frac{15}{16}$.
According to Dalton's law of partial pressure,the fraction of total pressure exerted by a gas is equal to its mole fraction.
Therefore,the fraction of total pressure exerted by hydrogen is $\frac{15}{16}$ or $15:16$.

(D) Let the weight of both ethane $(C_2H_6)$ and hydrogen $(H_2)$ be $W \ g$.
Number of moles of $C_2H_6$ $(n_{C_2H_6})$ = $\frac{W}{30}$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{W}{2}$.
According to Dalton's law of partial pressures, the fraction of total pressure exerted by a gas is equal to its mole fraction $(\chi)$.
$\chi_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{C_2H_6}} = \frac{\frac{W}{2}}{\frac{W}{2} + \frac{W}{30}}$.
$\chi_{H_2} = \frac{\frac{W}{2}}{\frac{15W + W}{30}} = \frac{W}{2} \times \frac{30}{16W} = \frac{15}{16}$.
Thus, the fraction of total pressure exerted by hydrogen is $\frac{15}{16}$ or $15: 16$.

(A) Given,mass of gas $(W) = 7.5 \ g$.
Volume of gas at $STP$ $(V) = 5.6 \ L$.
Since,moles of gas at $STP = \frac{V(L)}{22.4 \ L} = \frac{W}{M}$,where $M$ is the molar mass of the gas.
Therefore,$M = \frac{W \times 22.4}{V}$.
$M = \frac{7.5 \times 22.4}{5.6} = 30.00 \ g \ mol^{-1}$.
Among the given options,the molar mass $(M)$ of:
$(A) \ NO = 14 + 16 = 30.00 \ g \ mol^{-1}$.
$(B) \ N_{2}O = 28 + 16 = 44.00 \ g \ mol^{-1}$.
$(C) \ CO = 12 + 16 = 28.00 \ g \ mol^{-1}$.
$(D) \ CO_{2} = 12 + 32 = 44.00 \ g \ mol^{-1}$.
Since the molar mass of $NO$ is $30 \ g \ mol^{-1}$,the given gas is $NO$.
Hence,$A$ is the correct option.

(C) The molar mass of benzene $(C_6H_6)$ is $(6 \times 12) + (6 \times 1) = 78 \ g/mol$.
Number of moles of $C_6H_6 = \frac{39 \ g}{78 \ g/mol} = 0.5 \ mol$.
From the balanced chemical equation,$1 \ mol$ of $C_6H_6$ requires $\frac{15}{2} \ mol$ of $O_2$.
Therefore,$0.5 \ mol$ of $C_6H_6$ requires $\frac{15}{2} \times 0.5 = 3.75 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Volume of $O_2$ at $STP = 3.75 \ mol \times 22.4 \ L/mol = 84 \ L$.