Chemical stoichiometry Questions in English

(D) According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to their stoichiometric coefficients in the balanced chemical equation.
The balanced chemical equation is: $4NH_3(g) + 5O_2(g) \to 4NO(g) + 6H_2O(g)$.
From the stoichiometry,$5 \ mL$ of $O_2$ reacts with $4 \ mL$ of $NH_3$.
Therefore,$1 \ mL$ of $O_2$ reacts with $\frac{4}{5} \ mL$ of $NH_3$.
For $250 \ mL$ of $O_2$,the volume of $NH_3$ required is: $V_{NH_3} = \frac{4}{5} \times 250 \ mL = 200 \ mL$.

(A) Let the weight of both methane $(CH_4)$ and oxygen $(O_2)$ be $32\,g$.
Molar mass of $CH_4 = 16\,g/mol$.
Molar mass of $O_2 = 32\,g/mol$.
Moles of $CH_4$ $(n_{CH_4})$ = $\frac{32\,g}{16\,g/mol} = 2\,mol$.
Moles of $O_2$ $(n_{O_2})$ = $\frac{32\,g}{32\,g/mol} = 1\,mol$.
Total moles $(n_T)$ = $2 + 1 = 3\,mol$.
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 $(P_T)$.
Mole fraction of $O_2$ $(X_{O_2})$ = $\frac{n_{O_2}}{n_T} = \frac{1}{3}$.
Therefore,the fraction of the total pressure exerted by oxygen is $X_{O_2} = \frac{1}{3}$.

(D) The balanced chemical equation for the reaction is:
$Mg(s) + 2H_2O(l) \to Mg(OH)_2(aq) + H_2(g)$
The molar mass of $Mg$ is $24 \ g/mol$.
Number of moles of $Mg = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{24 \ g}{24 \ g/mol} = 1 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$.
Therefore,$1 \ mol$ of $H_2$ will be liberated.

(A) The balanced chemical equation for the decomposition of potassium chlorate is:
$2 KClO_3 \longrightarrow 2 KCl + 3 O_2$
From the stoichiometry of the reaction,$3 \, mol$ of $O_2$ ($3 \times 22.4 \, L$ at $STP$) is produced by $2 \, mol$ of $KClO_3$.
Therefore,the number of moles of $KClO_3$ required to produce $5.6 \, L$ of $O_2$ is:
$n(KClO_3) = \frac{2 \, mol \, KClO_3}{3 \times 22.4 \, L \, O_2} \times 5.6 \, L \, O_2$
$n(KClO_3) = \frac{2 \times 5.6}{67.2} = \frac{11.2}{67.2} = \frac{1}{6} \, mol$.

(D) The chemical reaction for the decomposition of limestone is: $CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)$
Mass of pure $CaCO_3 = 300 \ kg \times 0.90 = 270 \ kg = 270,000 \ g$.
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Moles of $CaCO_3 = \frac{270,000 \ g}{100 \ g/mol} = 2700 \ mol$.
According to the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CaO$.
Therefore,$2700 \ mol$ of $CaCO_3$ produces $2700 \ mol$ of $CaO$.
Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.
Weight of $CaO = 2700 \ mol \times 56 \ g/mol = 151,200 \ g = 151.2 \ kg$.

(D) Butane and isobutane are isomers and have the same molecular formula,$C_{4}H_{10}$.
The balanced chemical equation for the combustion of $C_{4}H_{10}$ is:
$C_{4}H_{10} + 6.5 O_{2} \to 4 CO_{2} + 5 H_{2}O$
Molar mass of $C_{4}H_{10} = (4 \times 12) + (10 \times 1) = 58 \, g/mol$.
From the stoichiometry,$58 \, g$ of $C_{4}H_{10}$ requires $6.5 \times 32 \, g$ of $O_{2}$.
Therefore,$1000 \, g$ $(1 \, kg)$ of $C_{4}H_{10}$ requires:
$\text{Mass of } O_{2} = \frac{6.5 \times 32 \times 1000}{58} \, g$
$\text{Mass of } O_{2} \approx 3586.2 \, g = 3.586 \, kg$.
Rounding to the given options,the correct answer is $3.58 \, kg$.

(C) The chemical reaction for the decomposition of limestone is:
$CaCO_{3(s)} \stackrel{\Delta}{\longrightarrow} CaO_{(s)} + CO_{2(g)}$
According to the stoichiometry,$100 \ g$ of $CaCO_3$ produces $44 \ g$ of $CO_2$.
Given mass of impure limestone = $10 \ kg$.
Mass of pure $CaCO_3$ = $80 \%$ of $10 \ kg = 0.80 \times 10 = 8 \ kg$.
Since $100 \ kg$ of $CaCO_3$ yields $44 \ kg$ of $CO_2$,
$8 \ kg$ of $CaCO_3$ will yield:
$\frac{44}{100} \times 8 = 3.52 \ kg$ of $CO_2$.

(B) The balanced chemical equation for the combustion of ethane is:
$2C_2H_6 + 7O_2 \longrightarrow 4CO_2 + 6H_2O$
Alternatively,using $1$ mole of $C_2H_6$:
$C_2H_6 + \frac{7}{2} O_2 \longrightarrow 2CO_2 + 3H_2O$
From the stoichiometry,$1$ mole of $C_2H_6$ $(30 \ g)$ produces $2$ moles of $CO_2$ ($2 \times 22.4 \ L$ at $STP$).
For $3 \ g$ of $C_2H_6$:
$\text{Moles of } C_2H_6 = \frac{3 \ g}{30 \ g/mol} = 0.1 \ mol$
$\text{Volume of } CO_2 = 0.1 \ mol \times 2 \times 22.4 \ L/mol = 4.48 \ L$.

(D) The rise in temperature is proportional to the heat evolved during the neutralization reaction. The heat evolved is maximum when the number of equivalents of acid and base are equal.
Let the volume of $H_2SO_4$ be $v \ mL$. Then the volume of $KOH$ is $(100 - v) \ mL$.
Normality of $H_2SO_4 = \text{Molarity} \times n\text{-factor} = 0.5 \times 2 = 1 \ N$.
Normality of $KOH = \text{Molarity} \times n\text{-factor} = 1 \times 1 = 1 \ N$.
For maximum neutralization,equivalents of acid = equivalents of base:
$v \times 1 = (100 - v) \times 1$
$v = 100 - v$
$2v = 100$
$v = 50 \ mL$.
Thus,volume of $H_2SO_4 = 50 \ mL$ and volume of $KOH = 50 \ mL$.

(A) Molar mass of $H_3PO_4 = (3 \times 1) + 31 + (4 \times 16) = 98 \ g/mol$.
Given mass of $H_3PO_4 = 98 \ g$,so moles of $H_3PO_4 = 98/98 = 1 \ mol$.
Each molecule of $H_3PO_4$ contains $3$ atoms of $H$.
Therefore,total moles of $H$ atoms $= 1 \ mol \times 3 = 3 \ mol$ of $H$ atoms.
Let the volume of $H_2$ gas at $STP$ be $V \ L$.
Number of moles of $H_2$ gas $= V/22.4 \ mol$.
Since each molecule of $H_2$ contains $2$ atoms of $H$,moles of $H$ atoms in $H_2$ gas $= 2 \times (V/22.4) = V/11.2 \ mol$.
Equating the moles of $H$ atoms: $V/11.2 = 3$.
$V = 3 \times 11.2 = 33.6 \ L$.

(B) Let the volume of each solution be $V \ L$.
Initial millimoles of $AgNO_3 = 0.1 \ M \times V \ L = 0.1V \ mmol$.
Since $AgNO_3$ dissociates completely as $AgNO_3 \rightarrow Ag^+ + NO_3^-$,the millimoles of $NO_3^-$ ions are equal to the millimoles of $AgNO_3$,which is $0.1V \ mmol$.
The total volume of the mixture $= V + V = 2V \ L$.
The concentration of $NO_3^-$ in the final mixture is given by:
$[NO_3^-] = \frac{\text{Total millimoles of } NO_3^-}{\text{Total volume}} = \frac{0.1V}{2V} = 0.05 \ M$.

(D) Let the total weight of the mixture be $100 \, g$.
Weight of $H_2 = 20 \, g$ and weight of $O_2 = 80 \, g$.
Moles of $H_2$ $(n_{H_2})$ = $\frac{20 \, g}{2 \, g/mol} = 10 \, mol$.
Moles of $O_2$ $(n_{O_2})$ = $\frac{80 \, g}{32 \, g/mol} = 2.5 \, mol$.
Mole fraction of $H_2$ $(x_{H_2})$ = $\frac{n_{H_2}}{n_{H_2} + n_{O_2}} = \frac{10}{10 + 2.5} = \frac{10}{12.5} = 0.8$.
Partial pressure of $H_2$ $(P_{H_2})$ = $x_{H_2} \times P_{Total} = 0.8 \times 1 \, atm = 0.8 \, atm$.

(B) The chemical formula for hydrated magnesium chloride is $MgCl_2 \cdot 6H_2O$.
When heated,it undergoes hydrolysis and thermal decomposition.
The reaction is: $MgCl_2 \cdot 6H_2O \xrightarrow{\Delta} MgO + 2HCl + 5H_2O$.
Thus,$5$ moles of $H_2O$ are liberated.

(C) Let the initial volume of $CO$ be $x \ mL$.
Therefore,the initial volume of $CO_2$ is $(240 - x) \ mL$.
The chemical reaction is: $CO_{2(g)} + C_{(s)} \longrightarrow 2CO_{(g)}$.
According to the reaction,$1 \ volume$ of $CO_2$ produces $2 \ volumes$ of $CO$.
So,$(240 - x) \ mL$ of $CO_2$ produces $2(240 - x) \ mL$ of $CO$.
The final volume of the mixture is the sum of the initial $CO$ and the $CO$ produced from $CO_2$:
$x + 2(240 - x) = 300$.
Solving for $x$:
$x + 480 - 2x = 300$.
$-x = 300 - 480$.
$-x = -180$.
$x = 180 \ mL$.

(A) Molecular mass $= 2 \times \text{Vapor density} = 2 \times 11.2 = 22.4 \, g \, mol^{-1}$.
At $STP$,the molar volume of any gas is $22.4 \, L \, mol^{-1}$.
Thus,$22.4 \, g$ of the gas occupies $22.4 \, L$.
Therefore,$1 \, g$ of the gas occupies $\frac{22.4 \, L}{22.4 \, g} = 1 \, L$.

(B) The balanced chemical equation is: $6CO_{2(g)} + 6H_2O_{(l)} \to C_6H_{12}O_{6(aq)} + 6O_{2(g)}$.
According to the stoichiometry,$6 \, moles$ of $CO_2$ produce $6 \, moles$ of $O_2$,which means $1 \, mole$ of $CO_2$ produces $1 \, mole$ of $O_2$.
At $STP$,$1 \, mole$ of any gas occupies $22400 \, mL$.
$112 \, mL$ of $O_2$ corresponds to $\frac{112}{22400} = 0.005 \, moles$ of $O_2$.
Since the molar ratio is $1:1$,$0.005 \, moles$ of $CO_2$ are required.
The molar mass of $CO_2$ is $44 \, g/mol$.
Mass of $CO_2 = 0.005 \, mol \times 44 \, g/mol = 0.22 \, g$.

(B) The balanced chemical equation for the combustion of carbon disulfide is:
$CS_2(l) + 3O_2(g) \longrightarrow CO_2(g) + 2SO_2(g)$
From the stoichiometry of the reaction,$1 \, \text{mol}$ of $CS_2$ requires $3 \, \text{mol}$ of $O_2$.
At $STP$,the volume of $1 \, \text{mol}$ of any ideal gas is $22.4 \, L$.
Therefore,the volume of $3 \, \text{mol}$ of $O_2$ is $3 \times 22.4 \, L = 67.2 \, L$.

(D) For a gas mixture,the weight $W$ is proportional to the product of volume $V$ and molar mass $M$ $(W \propto V \times M)$.
Let the total volume be $100 \ mL$. Then,$V_{He} = 50 \ mL$ and $V_{CH_4} = 50 \ mL$.
Molar mass of Helium $(M_1)$ = $4 \ g/mol$.
Molar mass of Methane $(M_2)$ = $16 \ g/mol$.
Weight of Helium $(W_1)$ = $50 \times 4 = 200$.
Weight of Methane $(W_2)$ = $50 \times 16 = 800$.
Percentage of Methane by weight = $\frac{W_2}{W_1 + W_2} \times 100 = \frac{800}{200 + 800} \times 100 = \frac{800}{1000} \times 100 = 80\%$.

(C) For the same temperature $T$ and pressure $P$,according to Avogadro's law,$n \propto V$.
$\therefore \frac{n_{N_2}}{n_{O_2}} = \frac{V_{N_2}}{V_{O_2}} = \frac{1}{7/8} = \frac{8}{7}$.
Since $n = \frac{m}{M}$,where $m$ is mass and $M$ is molar mass:
$\frac{m(N_2) / M(N_2)}{m(O_2) / M(O_2)} = \frac{8}{7}$.
Given $M(N_2) = 28 \, g/mol$ and $M(O_2) = 32 \, g/mol$:
$\frac{m(N_2) / 28}{m(O_2) / 32} = \frac{8}{7}$.
$\frac{m(N_2)}{m(O_2)} = \frac{8}{7} \times \frac{28}{32} = \frac{8}{7} \times \frac{7}{8} = 1$.
Therefore,$m(N_2) = m(O_2)$.

(D) At $STP$,the molar volume of an ideal gas is $22.4 \, L \, mol^{-1}$.
Using the formula: $\text{Density} (d) = \frac{\text{Molar Mass} (M)}{\text{Molar Volume} (V_m)}$.
$M = d \times V_m = 2.86 \, g \, L^{-1} \times 22.4 \, L \, mol^{-1} = 64.064 \, g \, mol^{-1}$.
The molar mass of $SO_2$ is $32 + (2 \times 16) = 64 \, g \, mol^{-1}$.
Therefore,the gas is $SO_2$.

(B) The chemical formula is $N_2O_5$.
The molar mass of $N_2$ is $2 \times 14 = 28 \ g$ and the molar mass of $O_5$ is $5 \times 16 = 80 \ g$.
According to the law of definite proportions,$80 \ g$ of oxygen combines with $28 \ g$ of nitrogen.
Therefore,$40 \ g$ of oxygen will combine with:
$\frac{28}{80} \times 40 = 14 \ g$ of nitrogen.

(C) $1$. Calculate the number of moles of $NaCl$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.85 \, g}{58.5 \, g/mol} = 0.1 \, mol$.
$2$. Calculate the molarity of the solution: $M = \frac{n}{V(L)} = \frac{0.1 \, mol}{1 \, L} = 0.1 \, M$.
$3$. In $1 \, mL$ $(10^{-3} \, L)$,the number of moles of $NaCl$ is $0.1 \times 10^{-3} = 10^{-4} \, mol$.
$4$. Each $NaCl$ unit dissociates into $2$ ions ($Na^+$ and $Cl^-$).
$5$. Total number of ions = $2 \times \text{moles} \times N_A = 2 \times 10^{-4} \times 6.022 \times 10^{23} = 1.2044 \times 10^{20}$ ions.

(B) The mass of $1 \, mol$ of iron is $55.85 \, g$.
Density is defined as $\text{mass} / \text{volume}$.
Therefore,$\text{volume} = \text{mass} / \text{density}$.
$\text{Volume} = \frac{55.85 \, g}{7.86 \, g \, cm^{-3}} = 7.11 \, cm^{3}$.

(C) The balanced chemical equation is: $SO_{2(g)} + 1/2 O_{2(g)} \to SO_{3(g)}$.
$1$ mole of $SO_2$ $(64 \, g)$ requires $1/2$ mole of $O_2$ ($0.5 \times 22.4 \, L = 11.2 \, L$ at $STP$).
For $6.4 \, g$ of $SO_2$,the volume of $O_2$ required is: $(6.4 \, g / 64 \, g/mol) \times 11.2 \, L/mol = 1.12 \, L$ of $O_2$.
Since air contains $20 \% \, O_2$ by volume,the volume of air required is: $(1.12 \, L / 20) \times 100 = 5.6 \, L$ of air.

(C) For $HCl$,normality $(N)$ is equal to molarity $(M)$ because the n-factor is $1$.
So,$0.5 \, N \, HCl = 0.5 \, M \, HCl$.
Number of moles of $HCl = \text{Molarity} \times \text{Volume in Liters} = 0.5 \times 0.5 = 0.25 \, \text{mol}$.
The balanced chemical equation is: $Zn + 2HCl \rightarrow ZnCl_2 + H_2$.
According to the stoichiometry,$2 \, \text{moles}$ of $HCl$ produce $1 \, \text{mole}$ of $H_2$ gas.
Therefore,$0.25 \, \text{moles}$ of $HCl$ will produce $\frac{0.25}{2} = 0.125 \, \text{moles}$ of $H_2$.
Volume of $H_2$ at $STP = \text{moles} \times 22.4 \, L = 0.125 \times 22.4 = 2.8 \, L$.

(C) The balanced chemical equation for the conversion of oxygen to ozone is:
$3O_{2(g)} \longrightarrow 2O_{3(g)}$
According to the stoichiometry,$2 \times 22.4 \, L$ of $O_3$ is produced by $3 \times 32 \, g$ of $O_2$.
To produce $33.6 \, L$ of $O_3$,the mass of $O_2$ required (assuming $100 \%$ conversion) is:
$Mass = \frac{3 \times 32 \times 33.6}{2 \times 22.4} = 72 \, g$
Since only $15 \%$ of $O_2$ is converted to $O_3$,the total mass of $O_2$ required is:
$Total \, mass = \frac{72 \times 100}{15} = 480 \, g$

(A) According to Avogadro's Law,under the same conditions of temperature and pressure,equal volumes of gases contain an equal number of moles $(n)$.
Mass $(m)$ = Number of moles $(n)$ $\times$ Molar mass $(M)$.
Since $n$ is constant for all three gases,the ratio of their masses is equal to the ratio of their molar masses.
Molar mass of $O_2 = 32 \ g/mol$.
Molar mass of $O_3 = 48 \ g/mol$.
Molar mass of $SO_2 = 64 \ g/mol$.
Ratio of masses = $32 : 48 : 64$.
Dividing by $16$,we get $2 : 3 : 4$.

(A) According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain an equal number of moles.
Since the volumes are equal ($3 \ L$ each),the number of moles of the hydrocarbon $(n_H)$ and $CO_2$ $(n_{CO_2})$ are equal.
$n_H = n_{CO_2}$
Given that the masses $(w)$ are also equal,the molar mass $(M)$ must be equal:
$\frac{w}{M_H} = \frac{w}{M_{CO_2}}$
$M_H = M_{CO_2}$
The molar mass of $CO_2 = 12 + (2 \times 16) = 44 \ g/mol$.
Therefore,the molar mass of the hydrocarbon is $44 \ g/mol$.
For $C_3H_8$: $M = (3 \times 12) + (8 \times 1) = 36 + 8 = 44 \ g/mol$.
Thus,the hydrocarbon is $C_3H_8$.

(C) The chemical reaction is: $CO_2(g) + C(s) \rightarrow 2CO(g)$.
Let the initial volume of $CO_2$ be $1 \ L$.
Let $x \ L$ be the volume of $CO_2$ that reacts.
According to the stoichiometry,$x \ L$ of $CO_2$ produces $2x \ L$ of $CO$.
The remaining volume of $CO_2$ is $(1 - x) \ L$.
The total volume of the gas mixture at $STP$ is $(1 - x) + 2x = 1 + x$.
Given that the total volume is $1.5 \ L$,we have $1 + x = 1.5$,so $x = 0.5 \ L$.
The volume of $CO$ produced is $2x = 2 \times 0.5 = 1 \ L$.
The number of moles of $CO$ at $STP$ is $\frac{\text{Volume}}{22.4} = \frac{1}{22.4}$ moles.

(B) The general combustion reaction for a hydrocarbon $C_xH_y$ is given by:
$C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O$
Comparing this with the general form,the coefficients are:
Coefficient of $O_2 = (x + y/4)$
Coefficient of $CO_2 = x$
Coefficient of $H_2O = y/2$
Thus,the correct option is $B$.

(B) The general combustion reaction for a hydrocarbon $C_xH_{2y}$ is:
$C_xH_{2y} + mO_2 \rightarrow xCO_2 + yH_2O$
To balance the oxygen atoms on both sides:
Total oxygen atoms on the right side = $(2 \times x) + (1 \times y) = 2x + y$.
Since there are $2m$ oxygen atoms on the left side:
$2m = 2x + y$
$m = x + y/2$
Therefore,the number of moles of $O_2$ required is $x + y/2$.

(A) The combustion reaction is represented as:
$C_xH_yO_z + mO_2 \longrightarrow xCO_2 + (y/2)H_2O$
To balance the oxygen atoms on both sides:
Total oxygen atoms on the left side = $z + 2m$
Total oxygen atoms on the right side = $2x + y/2$
Equating them:
$z + 2m = 2x + y/2$
$2m = 2x + y/2 - z$
$m = x + y/4 - z/2$

(A) The general combustion reaction for a hydrocarbon $C_xH_y$ is: $C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O$.
For the hydrocarbon $C_xH_{2n-2}$,we have $y = 2n - 2$.
Substituting $y$ into the coefficients:
Coefficient of $CO_2 = x$.
Coefficient of $H_2O = y/2 = (2n - 2)/2 = n - 1$.
Coefficient of $O_2 = x + y/4 = x + (2n - 2)/4 = x + (n - 1)/2$.
Thus,the coefficients are $x + (n - 1)/2, x, n - 1$.

(B) The balanced chemical equation is: $CaCO_3 + H_2SO_4 \longrightarrow CaSO_4 + H_2O + CO_2$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \, g/mol$.
Molar mass of $H_2SO_4 = (2 \times 1) + 32 + (4 \times 16) = 98 \, g/mol$.
From the stoichiometry,$100 \, g$ of $CaCO_3$ reacts with $98 \, g$ of pure $H_2SO_4$.
Since the $H_2SO_4$ solution is $60\%$,the mass of the solution required is:
$\text{Mass of solution} = \frac{\text{Mass of pure } H_2SO_4}{\text{Percentage}} \times 100 = \frac{98}{60} \times 100 = 163.33 \, g$.

(D) The chemical formula of the compound is $Cu_2O$.
The molar mass of $Cu$ is $63.5 \, g/mol$ and $O$ is $16 \, g/mol$.
In $Cu_2O$,$2 \times 63.5 = 127 \, g$ of $Cu$ is combined with $16 \, g$ of $O$.
Therefore,$127 \, g$ of $Cu$ corresponds to $16 \, g$ of $O$.
For $1.27 \, g$ of $Cu$,the amount of oxygen is calculated as:
$\text{Mass of } O = \frac{16 \, g \times 1.27 \, g}{127 \, g} = 0.16 \, g$.

(A) The balanced chemical equation is: $4Al + 3O_2 \longrightarrow 2Al_2O_3$.
From the stoichiometry,$3 \, mol$ of $O_2$ reacts with $4 \, mol$ of $Al$.
Therefore,$1.5 \, mol$ of $O_2$ will react with: $\frac{4}{3} \times 1.5 = 2 \, mol$ of $Al$.
The molar mass of $Al$ is $27 \, g/mol$.
Mass of $Al = 2 \, mol \times 27 \, g/mol = 54 \, g$.

(B) The balanced chemical equation is: $CH_4 + 2O_2 \to CO_2 + 2H_2O$.
From the stoichiometry of the reaction,$1 \, mol$ of $CH_4$ $(16 \, g)$ reacts with $2 \, mol$ of $O_2$.
Given mass of $CH_4 = 4.8 \, g$.
Moles of $CH_4 = \frac{4.8 \, g}{16 \, g/mol} = 0.3 \, mol$.
Since $1 \, mol$ of $CH_4$ requires $2 \, mol$ of $O_2$,then $0.3 \, mol$ of $CH_4$ requires: $0.3 \times 2 = 0.6 \, mol$ of $O_2$.

(B) From the first reaction: $1 \ mol$ of $S_8$ produces $8 \ mol$ of $SO_2$.
From the second reaction: $2 \ mol$ of $SO_2$ produces $2 \ mol$ of $SO_3$.
Therefore,$8 \ mol$ of $SO_2$ will produce $8 \ mol$ of $SO_3$.
The molar mass of $SO_3 = 32 + (3 \times 16) = 80 \ g/mol$.
Mass of $SO_3 = 8 \ mol \times 80 \ g/mol = 640 \ g$.

(D) For gases,the ratio of volumes is equal to the ratio of moles.
Let the hydrocarbon be $C_xH_y$.
The combustion reaction is: $C_xH_y + (x + \frac{y}{4}) O_2 \longrightarrow x CO_2 + \frac{y}{2} H_2O(g)$.
Given volumes: $0.5 \, L$ of $C_xH_y$ produces $2.5 \, L$ of $CO_2$ and $3.0 \, L$ of $H_2O$.
Dividing by $0.5 \, L$: $1 \, mol$ of $C_xH_y$ produces $5 \, mol$ of $CO_2$ and $6 \, mol$ of $H_2O$.
Comparing coefficients: $x = 5$ and $\frac{y}{2} = 6 \implies y = 12$.
Therefore,the molecular formula is $C_5H_{12}$.

(B) Given: Volume $(V) = 0.2 \, L = 200 \, mL$,Molarity $(M) = 0.2 \, M$.
Number of millimoles $= Molarity \times Volume \text{ (in } mL) = 0.2 \times 200 = 40 \, \text{mmol}$.
$Ba(OH)_2$ is a diacidic base,so its n-factor $= 2$.
Number of milliequivalents $= \text{millimoles} \times n\text{-factor} = 40 \times 2 = 80 \, \text{meq}$.
Thus,the values are $40$ and $80$ respectively.

(C) Step $1$: Calculate the molarity $(M_1)$ of the concentrated $H_2SO_4$ solution.
$M_1 = \frac{\% \, w/w \times d \times 10}{M_m} = \frac{96 \times 1.83 \times 10}{98} \approx 17.92 \, M \approx 18 \, M$.
Step $2$: Use the dilution formula $M_1V_1 = M_2V_2$.
Given: $M_2 = 3 \, M$,$V_2 = 2 \, L = 2000 \, mL$.
$V_1 = \frac{M_2 \times V_2}{M_1} = \frac{3 \times 2000}{18} = 333.33 \, mL$.

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

(D) Density of $HCl = 1.17 \, g \, cm^{-3}$.
Mass of $1 \, L$ (or $1000 \, mL$) of $HCl = 1.17 \, g \, mL^{-1} \times 1000 \, mL = 1170 \, g$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \, g \, mol^{-1}$.
Molarity $(M) = \frac{\text{Mass of solute in } g}{\text{Molar mass} \times \text{Volume of solution in } L}$.
$M = \frac{1170 \, g}{36.5 \, g \, mol^{-1} \times 1 \, L} = 32.05 \, M$.

(C) The molar mass of $Na_2CO_3$ is $106 \, g/mol$.
The molarity $(M_1)$ of the original solution is calculated as:
$M_1 = \frac{1000 \times \text{mass}}{M_m \times V(mL)} = \frac{1000 \times 2.65}{106 \times 250} = \frac{2650}{26500} = 0.1 \, M$.
Using the dilution formula $M_1 V_1 = M_2 V_2$:
Given $M_1 = 0.1 \, M$,$V_1 = 10 \, mL$,and $V_2 = 0.1 \, L = 100 \, mL$.
$M_2 = \frac{M_1 V_1}{V_2} = \frac{0.1 \times 10}{100} = 0.01 \, M$.

(D) The molar mass of $Na_2CO_3$ is $106 \, g/mol$.
The molarity of the initial solution is calculated as: $M_1 = \frac{\text{mass}}{\text{molar mass} \times \text{volume in L}} = \frac{5.3}{106 \times 0.250} = 0.2 \, M$.
Using the dilution formula $M_1 V_1 = M_2 V_2$:
$0.2 \, M \times 10 \, mL = M_2 \times 100 \, mL$.
$M_2 = \frac{0.2 \times 10}{100} = 0.02 \, M$.

(C) The molar mass of $Na_2CO_3$ is $M_m = (2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$.
The formula for Molarity $(Y)$ is $Y = \frac{X \times 1000}{M_m \times V(mL)}$.
Substituting the values: $Y = \frac{X \times 1000}{106 \times 100} = \frac{X}{10.6}$.
Checking option $(c)$: If $X = 1.06 \ g$,then $Y = \frac{1.06}{10.6} = 0.1 \ M$.
Thus,the correct values are $X = 1.06$ and $Y = 0.1$.

(B) The $NO_3^-$ ions are provided only by the $AgNO_3$ solution.
Since equal volumes of both solutions are mixed,the total volume becomes double the initial volume of each solution.
Let the volume of each solution be $V$. The total volume of the mixture is $2V$.
The number of moles of $NO_3^-$ is equal to the number of moles of $AgNO_3$,which is $0.1 \times V$.
The concentration of $NO_3^-$ in the mixture is given by:
$[NO_3^-] = \frac{\text{moles of } NO_3^-}{\text{total volume}} = \frac{0.1 \times V}{2V} = 0.05 \, M$.

(A) The normality formula is given by $N = \frac{W \times 1000}{E_q \times V(mL)}$.
For $HCl$,the equivalent weight $E_q$ is equal to the molar mass,which is $36.5 \, g/mol$.
Given: $N = 0.1 \, N$,$V = 1500 \, mL$,$E_q = 36.5 \, g/mol$.
Substituting the values: $0.1 = \frac{W \times 1000}{36.5 \times 1500}$.
$W = \frac{0.1 \times 36.5 \times 1500}{1000}$.
$W = 0.1 \times 36.5 \times 1.5$.
$W = 5.475 \, g$.