Chemical stoichiometry Questions in English

(C) According to Avogadro's hypothesis,the volume of a gas $(V)$ is directly proportional to the number of moles $(n)$ at constant temperature and pressure.
Therefore,the ratio of the volumes of gases is equal to the ratio of their moles.
Let the mass of each gas be $m$.
The molar masses are: $M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $M_{CH_4} = 16 \ g/mol$.
The ratio of volumes is:
$V_{H_2} : V_{O_2} : V_{CH_4} = n_{H_2} : n_{O_2} : n_{CH_4} = \frac{m}{2} : \frac{m}{32} : \frac{m}{16}$.
Multiplying by $32/m$,we get:
$16 : 1 : 2$.

(B) The reaction between the chloride and $AgNO_3$ is: $Ag^{+} + Cl^{-} \rightarrow AgCl_{(s)}$.
Number of moles of $AgNO_3 = \text{Molarity} \times \text{Volume (in L)} = 0.1 \ M \times 0.01 \ L = 10^{-3} \ mol$.
Number of moles of the chloride solution $= 0.05 \ M \times 0.01 \ L = 0.5 \times 10^{-3} \ mol$.
Let the formula of the chloride be $XCl_n$. The number of moles of $Cl^{-}$ ions provided by the solution is $n \times (0.5 \times 10^{-3}) \ mol$.
According to the stoichiometry of the reaction,moles of $Cl^{-}$ must equal moles of $Ag^{+}$:
$n \times 0.5 \times 10^{-3} = 10^{-3}$.
Solving for $n$: $n = \frac{10^{-3}}{0.5 \times 10^{-3}} = 2$.
Therefore,the formula of the chloride is $XCl_2$.

(A) The balanced chemical equation for the combustion of propane is:
$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$
According to Avogadro's Law,at constant temperature and pressure,the volume of gases reacting is directly proportional to their stoichiometric coefficients.
From the balanced equation,$1 \text{ volume of } C_3H_8$ requires $5 \text{ volumes of } O_2$.
Therefore,to burn $1 \ L$ of $C_3H_8$,the volume of $O_2$ required is $5 \times 1 \ L = 5 \ L$.

(B) The combustion reactions are:
$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$
$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$
Let the volume of $CH_4$ be $x \ L$ and the volume of $C_3H_8$ be $(5-x) \ L$.
According to the stoichiometry of the reactions,the volume of $O_2$ consumed is $2x + 5(5-x) = 16$.
$2x + 25 - 5x = 16 \implies 3x = 9 \implies x = 3 \ L$.
Thus,volume of $CH_4 = 3 \ L$ and volume of $C_3H_8 = 2 \ L$.
At $STP$ $(0 \ ^\circ C, 1 \ atm)$,$22.4 \ L$ corresponds to $1 \ mol$. Therefore,moles of $CH_4 = 3/22.4$ and moles of $C_3H_8 = 2/22.4$.
Heat released $= (3/22.4) \times 890 + (2/22.4) \times 2220 = (2670 + 4440) / 22.4 = 7110 / 22.4 \approx 317.4 \ kJ$.

(C) The decomposition reaction is: $MgCO_{3(s)} \longrightarrow MgO_{(s)} + CO_{2(g)}$
Molar mass of $MgCO_3 = 24 + 12 + (3 \times 16) = 84 \ g/mol$.
Molar mass of $MgO = 24 + 16 = 40 \ g/mol$.
According to the stoichiometry,$84 \ g$ of pure $MgCO_3$ produces $40 \ g$ of $MgO$.
Therefore,the mass of pure $MgCO_3$ required to produce $8.0 \ g$ of $MgO$ is: $\frac{84 \ g \ MgCO_3}{40 \ g \ MgO} \times 8.0 \ g \ MgO = 16.8 \ g \ MgCO_3$.
Percentage purity of the sample = $\frac{\text{Mass of pure } MgCO_3}{\text{Total mass of sample}} \times 100 = \frac{16.8}{20.0} \times 100 = 84 \ \%$.

(C) Step $1$: Calculate the number of moles of $HNO_3$ required.
$n = M \times V(L) = 2.0 \ mol/L \times 0.250 \ L = 0.5 \ mol$.
Step $2$: Calculate the mass of pure $HNO_3$ required.
Molar mass of $HNO_3 = 1 + 14 + (3 \times 16) = 63 \ g/mol$.
Mass of $HNO_3 = 0.5 \ mol \times 63 \ g/mol = 31.5 \ g$.
Step $3$: Calculate the mass of the $70\%$ concentrated solution.
Since the solution is $70\%$ $HNO_3$ by mass,$70 \ g$ of $HNO_3$ is present in $100 \ g$ of solution.
Mass of solution $= (100 \ g \text{ solution} / 70 \ g \text{ } HNO_3) \times 31.5 \ g \text{ } HNO_3 = 45 \ g$.

(C) First,calculate the molarity $(M_1)$ of the concentrated acid:
$M_1 = \frac{\text{density} \times 10 \times \% \text{ by mass}}{\text{molar mass}} = \frac{1.80 \times 10 \times 98}{98} = 18 \ M$.
Now,use the dilution formula $M_1 V_1 = M_2 V_2$:
$18 \times V_1 = 0.1 \times 1000 \ mL$.
$V_1 = \frac{100}{18} \approx 5.55 \ mL$.

(D) The relationship between Molarity $(M)$,percentage by mass $(\% \; w/w)$,and density $(d)$ is given by the formula:
$M = \frac{10 \times \% \; w/w \times d}{M_{solute}}$
Given values are:
$M = 3.60 \; mol \; L^{-1}$
$\% \; w/w = 29 \%$
$M_{solute} = 98 \; g \; mol^{-1}$
Substituting these values into the formula:
$3.60 = \frac{10 \times 29 \times d}{98}$
$d = \frac{3.60 \times 98}{10 \times 29}$
$d = \frac{352.8}{290} \approx 1.2165 \; g \; mL^{-1}$
Rounding to two decimal places,we get $d = 1.22 \; g \; mL^{-1}$.

(D) From the balanced chemical equation: $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_{2(g)}$
$6 \ moles$ of $HCl_{(aq)}$ produces $3 \ moles$ of $H_{2(g)}$.
Therefore,$1 \ mole$ of $HCl_{(aq)}$ produces $\frac{3}{6} = 0.5 \ moles$ of $H_{2(g)}$.
At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$.
Thus,$0.5 \ moles$ of $H_{2(g)}$ occupies $0.5 \times 22.4 \ L = 11.2 \ L$ at $STP$.
Hence,option $D$ is correct.

(B) The ion exchange reaction is: $2(C_8H_7SO_3^- Na^+) + Ca^{2+} \rightarrow (C_8H_7SO_3^-)_2 Ca^{2+} + 2Na^+$.
From the stoichiometry,$2 \ mol$ of the resin are required to exchange $1 \ mol$ of $Ca^{2+}$ ions.
Therefore,$1 \ mol$ of the resin can take up $\frac{1}{2} \ mol$ of $Ca^{2+}$ ions.
Given the molar mass of the resin is $206 \ g/mol$,the uptake capacity per gram of resin is $\frac{1 \ mol \ Ca^{2+}}{2 \times 206 \ g \ resin} = \frac{1}{412} \ mol/g$.

(D) The combustion reaction is: $C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O_{(l)}$
Volume of $O_2$ used $= 375 \times 0.20 = 75 \ mL$.
From the stoichiometry: $15(x + \frac{y}{4}) = 75$,which simplifies to $x + \frac{y}{4} = 5$,or $4x + y = 20$.
The final volume $330 \ mL$ consists of $CO_2$ produced and the remaining $N_2$ from the air.
Volume of $N_2 = 375 - 75 = 300 \ mL$.
Volume of $CO_2 = 330 - 300 = 30 \ mL$.
Since $15 \ mL$ of hydrocarbon produces $30 \ mL$ of $CO_2$,$x = 2$. Wait,checking stoichiometry: $15x = 30 \implies x = 2$.
Substituting $x=2$ into $4x + y = 20$: $4(2) + y = 20 \implies y = 12$. This suggests $C_2H_{12}$ which is impossible.
Re-evaluating: The total volume $330 \ mL$ includes $CO_2$ and $N_2$. $15x = 30 \implies x = 2$. If $x=3$,$15x = 45 \ mL$ of $CO_2$. Then $N_2 = 300 \ mL$,total $= 345 \ mL$. If $x=3$,$4(3) + y = 20 \implies y = 8$. This gives $C_3H_8$. $15 \ mL$ of $C_3H_8$ produces $45 \ mL$ of $CO_2$. Total volume $= 45 + 300 = 345 \ mL$. Given $330 \ mL$,let's check $C_3H_6$: $x=3, y=6$. $4(3)+6 = 18 \neq 20$. The correct hydrocarbon is $C_3H_8$.

(D) Let the mass of methane $(CH_4)$ and oxygen $(O_2)$ be $m \ g$.
Moles of $CH_4 = \frac{m}{16}$
Moles of $O_2 = \frac{m}{32}$
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 $O_2 (x_{O_2}) = \frac{\text{Moles of } O_2}{\text{Moles of } O_2 + \text{Moles of } CH_4}$
$x_{O_2} = \frac{m/32}{m/32 + m/16} = \frac{m/32}{m/32 + 2m/32} = \frac{m/32}{3m/32} = 1/3$
Therefore,the fraction of the total pressure exerted by oxygen is $1/3$.

(C) The mass of hydrogen in a $75 \ kg$ human body is calculated as: $\text{Mass of } H = \frac{10}{100} \times 75 \ kg = 7.5 \ kg$.
When all $^1H$ atoms (atomic mass $\approx 1$) are replaced by $^2H$ atoms (atomic mass $\approx 2$),the mass of the hydrogen component doubles.
New mass of hydrogen $= 7.5 \ kg \times 2 = 15 \ kg$.
Therefore,the weight gain is $15 \ kg - 7.5 \ kg = 7.5 \ kg$.

(B) The balanced chemical equation for the reaction is:
$M_{2}CO_{3} + 2HCl \rightarrow 2MCl + H_{2}O + CO_{2}$
According to the stoichiometry of the reaction,$1\, mol$ of $M_{2}CO_{3}$ produces $1\, mol$ of $CO_{2}$.
Therefore,the moles of $M_{2}CO_{3}$ reacted is equal to the moles of $CO_{2}$ produced,which is $0.01186\, mol$.
Using the formula: $\text{Molar mass} = \frac{\text{Mass}}{\text{Moles}}$
$\text{Molar mass} = \frac{1\, g}{0.01186\, mol} \approx 84.3\, g\, mol^{-1}$.

(B) Step $1$: Calculate moles of $FeS_2$. Molar mass of $FeS_2 = 56 + 2 \times 32 = 120 \ g/mol$. Moles of $FeS_2 = \frac{20 \ g}{120 \ g/mol} = \frac{1}{6} \ mol$.
Step $2$: Write the roasting reaction: $2FeS_2 + \frac{11}{2}O_2 \to Fe_2O_3 + 4SO_2$.
From stoichiometry,$2 \ mol$ of $FeS_2$ produce $4 \ mol$ of $SO_2$. Thus,$\frac{1}{6} \ mol$ of $FeS_2$ produces $\frac{1}{6} \times 2 = \frac{1}{3} \ mol$ of $SO_2$.
Step $3$: Write the absorption reaction: $SO_2 + 2NaOH \to Na_2SO_3 + H_2O$.
To react with $\frac{1}{3} \ mol$ of $SO_2$,we need $\frac{2}{3} \ mol$ of $NaOH$.
Step $4$: Calculate molarity. Given that $50\%$ of the $NaOH$ solution was used,let $M$ be the molarity. Total moles of $NaOH = \frac{400 \times M}{1000} = 0.4M$.
$0.5 \times 0.4M = \frac{2}{3} \ mol$ $\Rightarrow 0.2M = \frac{2}{3}$ $\Rightarrow M = \frac{2}{0.6} = \frac{10}{3} \ M$.

(D) The balanced chemical equation for the reaction between $H_2O_2$ and $KMnO_4$ in acidic medium is:
$2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$
From the stoichiometry,$2 \, \text{moles of } KMnO_4$ react with $5 \, \text{moles of } H_2O_2$.
Given: Volume of $KMnO_4 = 5 \, mL = 5 \times 10^{-3} \, L$,Molarity of $KMnO_4 = 1/25 \, M = 0.04 \, M$.
Moles of $KMnO_4 = \text{Molarity} \times \text{Volume} = 0.04 \times 5 \times 10^{-3} = 2 \times 10^{-4} \, \text{moles}$.
Using the stoichiometry,moles of $H_2O_2 = \frac{5}{2} \times (\text{moles of } KMnO_4) = 2.5 \times 2 \times 10^{-4} = 5 \times 10^{-4} \, \text{moles}$.
Therefore,the correct statement is that the moles of $H_2O_2$ in the solution $= 5 \times 10^{-4}$.

(A) $1$. Mass of krill consumed by the whale per day = $75 kg \times 10 = 750 kg = 7.5 \times 10^5 g$.
$2$. Mass of diatoms consumed by krill to produce this mass of krill = $750 kg \times 10 = 7500 kg = 7.5 \times 10^6 g$.
$3$. Assuming the mass of diatoms is equivalent to the mass of carbohydrates $(C_6H_{12}O_6)$ produced,moles of $C_6H_{12}O_6$ = $\frac{7.5 \times 10^6 g}{180 g/mol} = 41666.67 mol$.
$4$. From the balanced equation,$1 mol$ of $C_6H_{12}O_6$ requires $6 mol$ of $CO_2$.
$5$. Moles of $CO_2$ required = $41666.67 mol \times 6 = 250000 mol = 2.5 \times 10^5 mol$.

(NONE) Step $1$: Determine the normality of $KHC_2O_4$ solution.
Using the equivalence principle,$N_1V_1 = N_2V_2$.
$0.1 \ N \times 40 \ mL = N_2 \times 30 \ mL$.
$N_2 = \frac{4}{30} = \frac{2}{15} \ N$.
Step $2$: Titration of $KHC_2O_4$ with $KOH$.
$KHC_2O_4$ acts as an acid,reacting with $KOH$ in a $1:1$ molar ratio (or $1:1$ equivalent ratio).
$N_{acid} \times V_{acid} = N_{base} \times V_{base}$.
$\frac{2}{15} \ N \times 60 \ mL = 0.1 \ N \times V_{base}$.
$V_{base} = \frac{2 \times 60}{15 \times 0.1} = \frac{8}{0.1} = 80 \ mL$.

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Moles of oxalic acid in $400 \ mL$ = $\frac{18 \ g}{126 \ g/mol} = \frac{1}{7} \ mol$.
Moles of oxalic acid in $50 \ mL$ = $\frac{1}{7} \times \frac{50}{400} = \frac{1}{56} \ mol$.
The neutralization reaction is: $H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$.
According to the stoichiometry,$1 \ mol$ of $H_2C_2O_4$ reacts with $2 \ mol$ of $NaOH$.
Moles of $NaOH$ required = $2 \times \frac{1}{56} = \frac{1}{28} \ mol$.
Volume of $0.1 \ M \ NaOH$ = $\frac{\text{moles}}{\text{molarity}} = \frac{1/28}{0.1} = \frac{10}{28} \approx 0.357 \ L = 357 \ mL$.
Note: If anhydrous oxalic acid ($H_2C_2O_4$,molar mass $90 \ g/mol$) is assumed:
Moles in $400 \ mL$ = $\frac{18}{90} = 0.2 \ mol$.
Moles in $50 \ mL$ = $0.2 \times \frac{50}{400} = 0.025 \ mol$.
Moles of $NaOH$ required = $2 \times 0.025 = 0.05 \ mol$.
Volume of $0.1 \ M \ NaOH$ = $\frac{0.05}{0.1} = 0.5 \ L = 500 \ mL$.
Given the options,the calculation assumes anhydrous oxalic acid.

(B) Let the volume of $C_2H_6$ be $x \ L$ and the volume of $C_3H_8$ be $(3-x) \ L$.
The combustion reactions are:
$C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$
$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$
From the stoichiometry,$1 \ L$ of $C_2H_6$ produces $2 \ L$ of $CO_2$,and $1 \ L$ of $C_3H_8$ produces $3 \ L$ of $CO_2$.
Total volume of $CO_2 = 2(x) + 3(3-x) = 8$.
$2x + 9 - 3x = 8$.
$-x = -1$.
$x = 1 \ L$.

(D) The balanced chemical equations are:
$4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(l)}$
$4 NO_{(g)} + 2 O_{2(g)} \rightarrow 4 NO_{2(g)}$
Adding these equations,we get the overall reaction:
$4 NH_{3(g)} + 7 O_{2(g)} \rightarrow 4 NO_{2(g)} + 6 H_2O_{(l)}$
To obtain the maximum mass of $NO_2$,the reactants must be in their stoichiometric ratio to ensure there is no limiting reagent.
The molar ratio is $n_{NH_3} : n_{O_2} = 4 : 7$.
The mass ratio is calculated as:
$\text{Mass ratio} = (4 \times \text{molar mass of } NH_3) : (7 \times \text{molar mass of } O_2)$
$\text{Mass ratio} = (4 \times 17) : (7 \times 32)$
$\text{Mass ratio} = 68 : 224$
Dividing both sides by $4$,we get $17 : 56$.

(D) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and reduces to $Mn^{2+}$,involving a change in oxidation state of $+5$ ($n$-factor $= 5$).
$FeSO_4$ contains $Fe^{2+}$,which gets oxidized to $Fe^{3+}$ ($n$-factor $= 1$).
$Fe_2(SO_4)_3$ contains $Fe^{3+}$,which cannot be further oxidized by $KMnO_4$.
Equating the milliequivalents (meq) of the reducing agent $(FeSO_4)$ and the oxidizing agent $(KMnO_4)$:
$\text{meq of } FeSO_4 = \text{meq of } KMnO_4$
$\text{moles of } FeSO_4 \times 1 = \text{Molarity} \times \text{Volume (in mL)} \times n\text{-factor}$
$\text{moles of } FeSO_4 = 2 \times 100 \times 5 / 1000 = 1 \text{ mole}$.
Total moles of mixture $= 3 \text{ moles}$.
$\text{Mole fraction of } FeSO_4 = \frac{\text{moles of } FeSO_4}{\text{Total moles}} = \frac{1}{3}$.

(C) At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$2.24 \ L$ of the mixture corresponds to $\frac{2.24}{22.4} = 0.1 \ mol$ of the mixture.
Since the mixture is equimolecular,it contains equal moles of $CH_4$ and $C_2H_6$.
Moles of $CH_4 = 0.05 \ mol$ and moles of $C_2H_6 = 0.05 \ mol$.
Mass of $CH_4 = 0.05 \ mol \times 16 \ g/mol = 0.8 \ g$.
Mass of $C_2H_6 = 0.05 \ mol \times 30 \ g/mol = 1.5 \ g$.
Total mass of the mixture $= 0.8 \ g + 1.5 \ g = 2.3 \ g$.

(B) The reaction is: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$.
Number of millimoles of $NaOH = 50 \ cm^3 \times 0.4 \ M = 20 \ mmol$.
Since $2 \ mol$ of $NaOH$ react with $1 \ mol$ of $H_2SO_4$,the $20 \ mmol$ of $NaOH$ will require $10 \ mmol$ of $H_2SO_4$.
Volume of $H_2SO_4$ required $= \frac{10 \ mmol}{0.25 \ M} = 40 \ cm^3$.
As $H_2SO_4$ is added,the temperature of the solution increases due to the exothermic nature of the neutralisation reaction until the equivalence point is reached at $40 \ cm^3$.
After the equivalence point,the addition of excess $H_2SO_4$ (which is at the same initial temperature) causes the total volume of the solution to increase,leading to a decrease in temperature due to dilution.
Therefore,the temperature plot should show a maximum at $40 \ cm^3$. Plot $B$ shows the maximum temperature at $40 \ cm^3$.

(B) According to Dalton's law of partial pressure,$P_i = x_i \times P_{total}$,where $x_i$ is the mole fraction of the gas.
For the partial pressures to be equal $(P_{CO_2} = P_{N_2})$,their mole fractions must be equal $(x_{CO_2} = x_{N_2})$.
Since $x_i = \frac{n_i}{n_{total}}$,equal mole fractions imply that the number of moles must be equal $(n_{CO_2} = n_{N_2})$.
Let the number of moles be $n$.
Mass of $CO_2 = n \times \text{Molar mass of } CO_2 = n \times 44 \ g/mol$.
Mass of $N_2 = n \times \text{Molar mass of } N_2 = n \times 28 \ g/mol$.
The ratio of masses = $\frac{n \times 44}{n \times 28} = \frac{44}{28} = \frac{11}{7}$.

(C) The chemical equation for the reaction is: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$.
According to the stoichiometry,$1 \ mol$ of $CaCO_3$ $(100 \ g)$ produces $22400 \ mL$ of $CO_2$ at $STP$.
If the marble were $100 \%$ pure,$10 \ g$ of $CaCO_3$ would produce: $\frac{22400 \ mL}{100 \ g} \times 10 \ g = 2240 \ mL$ of $CO_2$.
However,only $1120 \ mL$ of $CO_2$ was obtained.
The percentage purity of $CaCO_3$ is calculated as: $\frac{\text{Actual volume of } CO_2}{\text{Theoretical volume of } CO_2} \times 100 = \frac{1120 \ mL}{2240 \ mL} \times 100 = 50 \%$.

(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by:
$C_xH_y + (x + y/4)O_2 \to xCO_2 + (y/2)H_2O$
According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to the number of moles.
Given:
Volume of $C_xH_y = 300 \ mL$
Volume of $CO_2 = 2.4 \ L = 2400 \ mL$
Volume of $H_2O = 2.7 \ L = 2700 \ mL$
From the stoichiometry:
$x = \text{Volume of } CO_2 / \text{Volume of } C_xH_y = 2400 / 300 = 8$
$y/2 = \text{Volume of } H_2O / \text{Volume of } C_xH_y = 2700 / 300 = 9$
$y = 9 \times 2 = 18$
Therefore,the molecular formula is $C_8H_{18}$.

(C) The reaction between $Na_2CO_3$ and $HCl$ is: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$.
The number of equivalents of $Na_2CO_3$ in $20 \ mL$ is equal to the number of equivalents of $HCl$ used.
Equivalents of $HCl = N \times V(L) = \frac{1}{10} \times \frac{19.8}{1000} = 1.98 \times 10^{-3} \ eq$.
Since $20 \ mL$ contains $1.98 \times 10^{-3} \ eq$,the total $100 \ mL$ contains $1.98 \times 10^{-3} \times \frac{100}{20} = 9.9 \times 10^{-3} \ eq$.
The equivalent weight of $Na_2CO_3$ is $\frac{106}{2} = 53 \ g/eq$.
Total mass of $Na_2CO_3$ (anhydrous) = $9.9 \times 10^{-3} \times 53 = 0.5247 \ g$.
The mass of $Na_2CO_3 \cdot xH_2O$ is $0.7 \ g$.
The fraction of $Na_2CO_3$ in the hydrate is $\frac{106}{106 + 18x} = \frac{0.5247}{0.7} \approx 0.7495$.
$106 = 0.7495 \times (106 + 18x) \implies 106 = 79.45 + 13.49x \implies 13.49x = 26.55 \implies x \approx 2$.

(C) The balanced chemical equation for the decomposition of phosphine is: $4PH_{3(g)} \rightarrow P_{4(g)} + 6H_{2(g)}$.
According to the stoichiometry,$4$ volumes of $PH_3$ produce $1 + 6 = 7$ volumes of gaseous products.
For $100 \ mL$ of $PH_3$:
Volume of $P_4 = \frac{1}{4} \times 100 = 25 \ mL$.
Volume of $H_2 = \frac{6}{4} \times 100 = 150 \ mL$.
Total final volume = $25 + 150 = 175 \ mL$.
Change in volume = $175 \ mL - 100 \ mL = +75 \ mL$.

(C) The balanced chemical equation is $2A + 3B + C \rightarrow 4D + 2E$.
First,calculate the mass of one stoichiometric unit of the reactants: $2(40) + 3(30) + 1(20) = 80 + 90 + 20 = 190 \ g$.
Since $570 \ g$ of the mixture is used,the number of stoichiometric units reacting is $570 / 190 = 3$.
According to the stoichiometry,$1$ unit of reaction produces $2$ moles of $E$. Therefore,$3$ units produce $3 \times 2 = 6 \ moles$ of $E$.
Using the law of conservation of mass: $Mass_{reactants} = Mass_{products}$.
$570 \ g = Mass_{D} + Mass_{E}$.
Mass of $D$ produced $= 3 \times (4 \times 15) = 3 \times 60 = 180 \ g$.
Mass of $E = 570 \ g - 180 \ g = 390 \ g$.

(C) The balanced chemical equation for the reaction is:
$4 NH_3 + 7 O_2 \longrightarrow 4 NO_2 + 6 H_2O$
To obtain the maximum product without any limiting reagent,the reactants must be in their stoichiometric ratio.
The molar ratio of $NH_3$ to $O_2$ is $4:7$.
The mass ratio is calculated as:
$\frac{\text{Mass of } NH_3}{\text{Mass of } O_2} = \frac{4 \times \text{Molar mass of } NH_3}{7 \times \text{Molar mass of } O_2}$
$= \frac{4 \times 17}{7 \times 32} = \frac{68}{224} = \frac{17}{56}$

(B) Step $1$: Calculate the moles of $NaOH$ to be neutralized.
Moles of $NaOH = \frac{100 \ g}{40 \ g/mol} = 2.5 \ mol$.
Step $2$: Determine the moles of $H_2SO_4$ required.
The neutralization reaction is $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
From the stoichiometry,$1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$.
So,moles of $H_2SO_4$ required $= \frac{2.5}{2} = 1.25 \ mol$.
Step $3$: Determine the moles of $S$ atoms required.
Since $1 \ mol$ of $H_2SO_4$ contains $1 \ mol$ of $S$ atoms,$1.25 \ mol$ of $H_2SO_4$ requires $1.25 \ mol$ of $S$ atoms.
Step $4$: Calculate the molecular weight of the solid.
The solid contains $20\%$ by mole of $S$ atoms.
Let the total moles of the solid be $n$.
Moles of $S = 0.20 \times n = 1.25 \ mol$.
Therefore,$n = \frac{1.25}{0.20} = 6.25 \ mol$.
Molecular weight of the solid $= \frac{\text{Total mass}}{\text{Total moles}} = \frac{400 \ g}{6.25 \ mol} = 64 \ g/mol$.

(B) The balanced chemical equation for the reaction is:
$PCl_{5(s)} + H_2O_{(g)} \rightarrow POCl_{3(s)} + 2HCl_{(g)}$
Molar mass of $PCl_5 = 31 + 5 \times 35.5 = 208.5 \ g/mol$.
Molar mass of $POCl_3 = 31 + 16 + 3 \times 35.5 = 153.5 \ g/mol$.
Let $x$ moles of $PCl_5$ react.
Initial mass of $PCl_5 = 417 \ g$.
Initial moles of $PCl_5 = 417 / 208.5 = 2 \ mol$.
According to the reaction,$x$ moles of $PCl_5$ produce $x$ moles of $POCl_3$ and $2x$ moles of $HCl_{(g)}$.
Since the container is open,$HCl_{(g)}$ escapes.
The solid remaining consists of $(2 - x)$ moles of $PCl_5$ and $x$ moles of $POCl_3$.
Mass of solid remaining $= (2 - x) \times 208.5 + x \times 153.5 = 395$.
$417 - 208.5x + 153.5x = 395$.
$417 - 55x = 395$.
$55x = 22$.
$x = 22 / 55 = 0.4 \ mol$.
Percentage conversion $= (x / 2) \times 100 = (0.4 / 2) \times 100 = 20 \%$.

(C) Total hardness $= 200 \ ppm$ of $Mg^{2+}$.
Temporary hardness is due to $Mg(HCO_3)_2$ and permanent hardness is due to $MgCl_2$.
$Na_2CO_3$ removes temporary hardness by precipitating $Mg^{2+}$ as $MgCO_3$.
Remaining hardness is due to $MgCl_2$,which is $\frac{1000}{3} \ ppm$ of $CaCO_3$ equivalent.
$1 \ ppm = 1 \ mg/L$. So,remaining hardness $= \frac{1000}{3} \ mg/L$ of $CaCO_3$.
In $6 \ L$,total $CaCO_3$ equivalent $= \frac{1000}{3} \times 6 = 2000 \ mg = 2 \ g$ of $CaCO_3$.
Moles of $CaCO_3 = \frac{2 \ g}{100 \ g/mol} = 0.02 \ mol$.
Since $1 \ mol$ of $MgCl_2$ is equivalent to $1 \ mol$ of $CaCO_3$,moles of $MgCl_2 = 0.02 \ mol$.
Mass of $MgCl_2 = 0.02 \ mol \times 95 \ g/mol = 1.9 \ g$.

(D) The condition for mixing solutions is given by the equation: $N_{1}V_{1} + N_{2}V_{2} + N_{3}V_{3} = N_{R}(V_{1} + V_{2} + V_{3})$.
Given $N_{1} = 12$,$N_{2} = 6$,$N_{3} = 2$,and the resultant normality $N_{R} = 4$.
Substituting the values: $12V_{1} + 6V_{2} + 2V_{3} = 4(V_{1} + V_{2} + V_{3})$.
Simplifying: $12V_{1} + 6V_{2} + 2V_{3} = 4V_{1} + 4V_{2} + 4V_{3}$.
$8V_{1} + 2V_{2} = 2V_{3}$,which simplifies to $4V_{1} + V_{2} = V_{3}$.
Checking the options:
For $A$ $(1:1:5)$: $4(1) + 1 = 5$. This is correct.
For $B$ $(1:2:6)$: $4(1) + 2 = 6$. This is correct.
For $C$ $(2:1:9)$: $4(2) + 1 = 9$. This is correct.
For $D$ $(1:2:4)$: $4(1) + 2 = 6 \neq 4$. This is incorrect.

(D) The balanced chemical equation for the reaction between $KMnO_4$ and $HCl$ is:
$2 KMnO_4 + 16 HCl \rightarrow 2 MnCl_2 + 5 Cl_2 + 2 KCl + 8 H_2O$
Dividing the entire equation by $2$ to normalize per mole of $KMnO_4$:
$KMnO_4 + 8 HCl \rightarrow MnCl_2 + \frac{5}{2} Cl_2 + KCl + 4 H_2O$
From the stoichiometry,we see that $\frac{5}{2}$ moles of $Cl_2$ are produced along with $4$ moles of $H_2O$.
To find the moles of $H_2O$ per mole of $Cl_2$,we calculate:
$\text{Ratio} = \frac{4 \text{ moles } H_2O}{5/2 \text{ moles } Cl_2} = 4 \times \frac{2}{5} = \frac{8}{5}$ moles of $H_2O$.

(B) The number of equivalents of acid equals the number of equivalents of base for complete neutralization.
$Eqv. \text{ of } H_3PO_x = Eqv. \text{ of } NaOH$
$1 \, mole \times \text{Basicity} = \frac{\text{mass}}{\text{Equivalent mass of } NaOH} = \frac{80 \, g}{40 \, g/mol} = 2 \, Eqv.$
Thus,the Basicity of the acid is $2$.
For phosphorus oxoacids,$H_3PO_3$ has a basicity of $2$ (where $x = 3$).
Therefore,$x = 3$ and the acid is dibasic.

(B) The balanced chemical equation for the combustion of ethane is:
$C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O$
Calculate the moles of $C_2H_6$:
$mol \text{ of } C_2H_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{3\, g}{30\, g/mol} = 0.1\, mol$
From the stoichiometry,$1\, mol$ of $C_2H_6$ requires $3.5\, mol$ of $O_2$:
$mol \text{ of } O_2 = \frac{7}{2} \times 0.1 = 0.35\, mol$
Calculate the volume of $O_2$ at $STP$ $(1\, mol = 22.4\, L)$:
$Vol. \text{ of } O_2 = 0.35\, mol \times 22.4\, L/mol = 7.84\, L$

(B) From the balanced chemical equation: $2CHI_3 + 6Ag \to 6AgI_{(s)} + C_2H_{2(g)}$
$2$ moles of $CHI_3$ produce $1$ mole of $C_2H_2$.
Therefore,$0.01$ moles of $CHI_3$ will produce $\frac{1}{2} \times 0.01 = 0.005$ moles of $C_2H_2$.
At $NTP$,$1$ mole of any gas occupies $22400 \ mL$.
Volume of $C_2H_2 = 0.005 \times 22400 = 112 \ mL$.

(A) The balanced chemical equation for the reaction is:
$H_3PO_4 + 3NaOH \rightarrow Na_3PO_4 + 3H_2O$
$1$. Calculate the molar mass of orthophosphoric acid $(H_3PO_4)$:
$M(H_3PO_4) = (3 \times 1) + 31 + (4 \times 16) = 98 \ g/mol$.
$2$. Calculate the moles of $H_3PO_4$:
$n(H_3PO_4) = \frac{0.98 \ g}{98 \ g/mol} = 0.01 \ mol$.
$3$. From the stoichiometry,$1 \ mol$ of $H_3PO_4$ reacts with $3 \ mol$ of $NaOH$.
Therefore,$0.01 \ mol$ of $H_3PO_4$ reacts with $0.01 \times 3 = 0.03 \ mol$ of $NaOH$.
$4$. Calculate the molarity $(M)$ of $NaOH$ solution:
$M = \frac{n(NaOH)}{V(L)} = \frac{0.03 \ mol}{0.060 \ L} = 0.5 \ M$.

(C) The balanced chemical equation for the decomposition of $KClO_3$ is:
$2KClO_3 \rightarrow 2KCl + 3O_2$
From the stoichiometry,$2 \ mol$ of $KClO_3$ produces $3 \ mol$ of $O_2$.
Given volume of $O_2 = 67.2 \ L$ at $STP$ $(0 \ ^\circ C, 1 \ atm)$.
Since $1 \ mol$ of gas occupies $22.4 \ L$ at $STP$,
Moles of $O_2 = \frac{67.2 \ L}{22.4 \ L/mol} = 3 \ mol$.
Using the stoichiometry,$3 \ mol$ of $O_2$ is produced by $2 \ mol$ of $KClO_3$.
So,$2 \ mol$ of $KClO_3$ decomposed.
Given that only $50\%$ of the initial mass of $KClO_3$ decomposed,let the initial mass be $m \ g$.
Moles of $KClO_3$ decomposed = $\frac{0.5 \times m}{122.5 \ g/mol} = 2 \ mol$.
$0.5 \times m = 2 \times 122.5 = 245$.
$m = \frac{245}{0.5} = 490 \ g$.

(B) The reaction between $Al_2(SO_4)_3$ and $BaCl_2$ is: $Al_2(SO_4)_3 + 3BaCl_2 \rightarrow 2AlCl_3 + 3BaSO_4(s)$.
Initial moles of $BaCl_2 = 20 \ mL \times 6.6 \ M = 132 \ mmol$.
Initial moles of $Al_2(SO_4)_3 = 20 \ mL \times 0.2 \ M = 4 \ mmol$.
Since $BaCl_2$ is in excess,all $Al_2(SO_4)_3$ reacts with $12 \ mmol$ of $BaCl_2$ to form $8 \ mmol$ of $AlCl_3$ and $12 \ mmol$ of $BaSO_4$ precipitate.
Remaining moles of $BaCl_2 = 132 \ mmol - 12 \ mmol = 120 \ mmol$.
Total moles of $Cl^{-}$ ions = (moles from remaining $BaCl_2$) + (moles from $AlCl_3$ formed) = $(120 \times 2) + (8 \times 3) = 240 + 24 = 264 \ mmol$.
Total volume of solution = $20 \ mL + 20 \ mL = 40 \ mL$.
Concentration of $Cl^{-} = \frac{264 \ mmol}{40 \ mL} = 6.6 \ M$.

(B) The balanced chemical equation for the combustion of ethene $(C_2H_4)$ is:
$C_2H_4 + 3O_2 \to 2CO_2 + 2H_2O$
Calculate the number of moles of ethene:
$n_{C_2H_4} = \frac{\text{mass}}{\text{molar mass}} = \frac{560 \ g}{28 \ g/mol} = 20 \ mol$
From the stoichiometry,$1 \ mol$ of $C_2H_4$ requires $3 \ mol$ of $O_2$.
Therefore,$20 \ mol$ of $C_2H_4$ requires:
$n_{O_2} = 3 \times 20 \ mol = 60 \ mol$
Calculate the mass of oxygen:
$w_{O_2} = n_{O_2} \times \text{molar mass of } O_2 = 60 \ mol \times 32 \ g/mol = 1920 \ g$
Convert to $kg$:
$1920 \ g = 1.92 \ kg$

(D) The molar mass of anhydrous $Na_2SO_4$ is $(2 \times 23) + 32 + (4 \times 16) = 142 \, g/mol$.
The molar mass of water is $18 \, g/mol$.
The loss in weight corresponds to the mass of $n$ water molecules lost,which is $18n$.
The percentage loss in weight is given by: $\frac{\text{Mass of water}}{\text{Total mass of hydrated salt}} \times 100 = 55.9 \%$.
$\frac{18n}{142 + 18n} \times 100 = 55.9$.
$1800n = 55.9(142 + 18n)$.
$1800n = 7937.8 + 1006.2n$.
$793.8n = 7937.8$.
$n = \frac{7937.8}{793.8} \approx 10$.
Therefore,the value of $n$ is $10$.

(C) According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to the number of moles $(V \propto n)$.
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 volumes:
$V_{hydrocarbon} = 0.5 \ L$
$V_{CO_2} = 3.5 \ L$
$V_{H_2O} = 4 \ L$
From the stoichiometry:
$x = V_{CO_2} / V_{hydrocarbon} = 3.5 / 0.5 = 7$
$y/2 = V_{H_2O} / V_{hydrocarbon} = 4 / 0.5 = 8 \Rightarrow y = 16$
Thus,the molecular formula is $C_7H_{16}$.

(B) The balanced chemical equation for the combustion of acetylene $(C_2H_2)$ is:
$2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(l)$
According to the stoichiometry of the reaction,$2 \, \text{volumes}$ of $C_2H_2$ require $5 \, \text{volumes}$ of $O_2$ for complete combustion.
Therefore,for $2 \, \text{volumes}$ of acetylene,$5 \, \text{volumes}$ of dioxygen are required.

(B) The combustion reaction for ethene is: $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$
According to the stoichiometry,$1 \ L$ of $C_2H_4$ requires $3 \ L$ of $O_2$.
Therefore,$10 \ L$ of $C_2H_4$ requires $10 \times 3 = 30 \ L$ of $O_2$.
Since air contains $20\%$ oxygen by volume,the volume of air required is:
$V_{air} = \frac{V_{O_2}}{0.20} = \frac{30}{0.20} = 150 \ L$.

(C) Let the volume of $C_2H_6$ be $x \, L$ and the volume of $C_3H_8$ be $(5-x) \, L$.
The combustion reactions are:
$C_2H_6 + \frac{7}{2}O_2 \to 2CO_2 + 3H_2O$
$C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O$
From the stoichiometry,$x \, L$ of $C_2H_6$ produces $2x \, L$ of $CO_2$.
$(5-x) \, L$ of $C_3H_8$ produces $3(5-x) \, L$ of $CO_2$.
Total volume of $CO_2$ produced is $11 \, L$:
$2x + 3(5-x) = 11$
$2x + 15 - 3x = 11$
$-x = -4$
$x = 4 \, L$
Volume percentage of $C_2H_6 = \frac{\text{Volume of } C_2H_6}{\text{Total volume}} \times 100 = \frac{4}{5} \times 100 = 80 \%$.

(D) The balanced chemical equation for the thermal decomposition of potassium bicarbonate is:
$2KHCO_{3(s)} \to K_2CO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$
Molar mass of $KHCO_3 = 39 + 1 + 12 + (3 \times 16) = 100 \ g/mol$.
Number of moles of $KHCO_3 = \frac{20 \ g}{100 \ g/mol} = 0.2 \ mol$.
From the stoichiometry of the reaction,$2 \ mol$ of $KHCO_3$ produces $1 \ mol$ of $CO_2$.
Therefore,$0.2 \ mol$ of $KHCO_3$ produces $\frac{0.2}{2} = 0.1 \ mol$ of $CO_2$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Volume of $CO_2 = 0.1 \ mol \times 22.4 \ L/mol = 2.24 \ L$.

(C) The balanced chemical equation is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
According to the law of equivalence,the number of equivalents of $KMnO_4$ reacting is equal to the number of equivalents of $O_2$ produced.
Equivalents of $KMnO_4 = \text{Normality} \times \text{Volume (in } L) = 0.5 \ N \times 0.1 \ L = 0.05 \ eq$.
Since the equivalents of $O_2$ produced must also be $0.05$,and the equivalent weight of $O_2$ in this reaction (where $O^{-2} \to O^0$,$n$-factor $= 4$) is $\frac{32}{4} = 8 \ g/eq$:
Mass of $O_2 = 0.05 \ eq \times 8 \ g/eq = 0.4 \ g$.
At standard conditions $(STP)$,$32 \ g$ of $O_2$ occupies $22.4 \ L$.
Volume of $0.4 \ g$ of $O_2 = \frac{22.4 \ L}{32 \ g} \times 0.4 \ g = 0.28 \ L$.