Chemical stoichiometry Questions in English

(D) The reaction is: $4(CH_3)_2SiCl_2 + 4H_2O \rightarrow ((CH_3)_2SiO)_4 + 8HCl$.
First,calculate the molar mass of dimethyldichlorosilane $(CH_3)_2SiCl_2$: $2(12) + 6(1) + 28 + 2(35.5) = 24 + 6 + 28 + 71 = 129 \ g \ mol^{-1}$.
Number of moles of $(CH_3)_2SiCl_2 = \frac{516 \ g}{129 \ g \ mol^{-1}} = 4 \ mol$.
According to the stoichiometry,$4 \ mol$ of $(CH_3)_2SiCl_2$ produces $1 \ mol$ of the tetrameric cyclic product $X$ $(((CH_3)_2SiO)_4)$.
Molar mass of $X = 8(12) + 24(1) + 4(28) + 4(16) = 96 + 24 + 112 + 64 = 296 \ g \ mol^{-1}$.
Theoretical yield of $X = 1 \ mol \times 296 \ g \ mol^{-1} = 296 \ g$.
Given the percentage yield is $75 \%$,the actual weight of $X$ obtained = $296 \ g \times \frac{75}{100} = 222 \ g$.

(D) First,calculate the molarity of the stock solution:
$M = \frac{\text{density} \times 10 \times \% (w/w)}{\text{molar mass}} = \frac{1.25 \times 10 \times 29.2}{36.5} = 10 \ M$.
Now,use the dilution formula $M_1V_1 = M_2V_2$:
$10 \ M \times V_1 = 0.4 \ M \times 200 \ mL$.
$V_1 = \frac{0.4 \times 200}{10} = 8 \ mL$.

(C) The chemical reaction for the self-reduction process of Galena $(PbS)$ is:
$PbS + O_2 \rightarrow Pb + SO_2$
From the stoichiometry of the balanced equation,$1 \ mol$ of $O_2$ produces $1 \ mol$ of $Pb$.
Therefore,the number of moles of $O_2$ consumed equals the number of moles of $Pb$ produced:
$n_{O_2} = n_{Pb}$
Using the relation $n = \frac{w}{M}$,we get:
$\frac{w_{O_2}}{M_{O_2}} = \frac{w_{Pb}}{M_{Pb}}$
Given $M_{O_2} = 2 \times 16 = 32 \ g \ mol^{-1}$ and $M_{Pb} = 207 \ g \ mol^{-1}$.
For $w_{O_2} = 1 \ kg$:
$w_{Pb} = \frac{w_{O_2} \times M_{Pb}}{M_{O_2}} = \frac{1 \ kg \times 207}{32} = 6.46875 \ kg \approx 6.47 \ kg$.

(C) The oxidation of rhombic sulphur $(S_8)$ by concentrated $HNO_3$ produces sulphuric acid $(H_2SO_4)$ as the product with the highest oxidation state of sulphur $(+6)$.
The balanced chemical equation is:
$S_8 + 48 \ HNO_3 \longrightarrow 8 \ H_2SO_4 + 48 \ NO_2 + 16 \ H_2O$
From the stoichiometry of the reaction,$1$ mole of $S_8$ produces $16$ moles of $H_2O$.
Since $1$ mole of rhombic sulphur is $S_8$,we must consider the reaction for $1$ mole of $S_8$ atoms (as implied by the question context of $1$ mole of sulphur atoms,or if $1$ mole of $S_8$ is meant,the stoichiometry adjusts accordingly. Given the standard interpretation of $1$ mole of $S$ atoms):
For $1$ mole of $S$ atoms,the equation is: $S + 6 \ HNO_3 \longrightarrow H_2SO_4 + 6 \ NO_2 + 2 \ H_2O$.
Thus,$1$ mole of $S$ produces $2$ moles of $H_2O$.
Mass of $H_2O = 2 \ mol \times 18 \ g \ mol^{-1} = 36 \ g$.
However,if the question refers to $1$ mole of $S_8$ molecules,then $16$ moles of $H_2O$ are produced.
Mass of $H_2O = 16 \ mol \times 18 \ g \ mol^{-1} = 288 \ g$.

(D) The chemical reaction is: $3Sn + 6HCl + C_6H_5NO_2 \rightarrow C_6H_5NH_3^+Cl^- + 3SnCl_2 + 2H_2O$.
The molar mass of the organic salt (anilinium chloride,$C_6H_5NH_3Cl$) is $129 \ g/mol$.
Given that $1.29 \ g$ of the organic salt is produced,the number of moles of salt formed is $n = \frac{1.29 \ g}{129 \ g/mol} = 0.01 \ mol$.
From the stoichiometry of the reaction:
$1 \ mol$ of nitrobenzene produces $1 \ mol$ of organic salt.
$3 \ mol$ of $Sn$ are required to produce $1 \ mol$ of organic salt.
Therefore,for $0.01 \ mol$ of salt:
Amount of nitrobenzene $(y)$ $= 0.01 \ mol \times 123 \ g/mol = 1.23 \ g$.
Amount of $Sn$ $(x)$ $= 0.03 \ mol \times 119 \ g/mol = 3.57 \ g$.
Thus,$x = 3.57$ and $y = 1.23$.

(D) The balanced chemical equation for the reaction is:
$5[Fe(C_2O_4)_2(H_2O)_2]^{2-} + 3MnO_4^- + 24H^{+} \longrightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 22H_2O$
From the stoichiometry of the reaction,the rate of change of concentration is related to the stoichiometric coefficients as:
$-\frac{1}{24} \frac{d[H^{+}]}{dt} = -\frac{1}{3} \frac{d[MnO_4^-]}{dt}$
Therefore,the ratio of the rate of change of $[H^{+}]$ to the rate of change of $[MnO_4^-]$ is:
$\frac{d[H^{+}]}{dt} / \frac{d[MnO_4^-]}{dt} = \frac{24}{3} = 8$.

(B) The balanced chemical equation for the reaction of diborane with methanol is:
$B_2H_6 + 6 CH_3OH \longrightarrow 2 B(OCH_3)_3 + 6 H_2$
From the stoichiometry of the reaction,$1$ mole of $B_2H_6$ produces $2$ moles of the boron-containing product,trimethyl borate $(B(OCH_3)_3)$.
Therefore,$3$ moles of $B_2H_6$ will produce $3 \times 2 = 6$ moles of $B(OCH_3)_3$.

(B) The reaction of white phosphorus $(P_4)$ with $NaOH$ is:
$P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$
Moles of $P_4 = \frac{1.24 \ g}{124 \ g/mol} = 0.01 \ mol$.
From the stoichiometry,$1 \ mol$ of $P_4$ produces $1 \ mol$ of $PH_3$ (gas $Q$).
So,moles of $PH_3 = 0.01 \ mol$.
The reaction of $PH_3$ with $CuSO_4$ is:
$2PH_3 + 3CuSO_4 \longrightarrow Cu_3P_2 + 3H_2SO_4$
Moles of $CuSO_4$ required = $\frac{3}{2} \times \text{moles of } PH_3 = \frac{3}{2} \times 0.01 = 0.015 \ mol$.
Molar mass of $CuSO_4 = 63 + 32 + (4 \times 16) = 159 \ g/mol$.
Mass of $CuSO_4 = 0.015 \ mol \times 159 \ g/mol = 2.385 \ g$.
Rounding to two decimal places,the value is $2.39 \ g$.

(D) The total charge of the oxide $M_X Y_2 O_4$ must be zero.
Given that $Y$ is in $+3$ oxidation state and $O$ is in $-2$ oxidation state,the total negative charge is $4 \times (-2) = -8$.
The total positive charge from $Y$ is $2 \times (+3) = +6$.
Therefore,the total positive charge from $M$ must be $8 - 6 = +2$.
Let the total number of $M$ ions be $X$. The fraction of $M^{2+}$ is $\frac{1}{3}$,so the number of $M^{2+}$ ions is $\frac{X}{3}$ and the number of $M^{3+}$ ions is $\frac{2X}{3}$.
The total charge contributed by $M$ is $(\frac{X}{3} \times 2) + (\frac{2X}{3} \times 3) = \frac{2X}{3} + 2X = \frac{8X}{3}$.
Equating the total positive charge from $M$ to $+2$,we get $\frac{8X}{3} = 2$.
Solving for $X$,we get $X = \frac{6}{8} = 0.75$.

(A) The reaction between $CO_2$ and $Ca(OH)_2$ is: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$.
Let the moles of $CO_2$ be $n$.
According to the stoichiometry,$n$ moles of $CO_2$ react with $n$ moles of $Ca(OH)_2$.
Given that this amount of $CO_2$ reacted with exactly half the initial amount of $Ca(OH)_2$,the total initial moles of $Ca(OH)_2 = 2n$.
Excess $Ca(OH)_2 = 2n - n = n$.
This excess $Ca(OH)_2$ is neutralized by $HCl$: $Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$.
Equivalents of $Ca(OH)_2 = $ Equivalents of $HCl$.
$n \times 2 = 0.1 \times 0.040 \times 1$.
$2n = 0.004 \implies n = 0.002 \ mol$.
At $STP$ $(1 \ atm, 273 \ K)$,$1 \ mol$ of gas occupies $22400 \ cm^3$.
Volume $x = 0.002 \times 22400 = 44.8 \ cm^3$.
Rounding to the nearest integer,$x = 45 \ cm^3$.

(D) Let the initial pressure of $N_2O_5$ be $P_0$.
The reaction is: $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$
At $t=0$: $P_0$,$0$,$0$
At $t=t$: $P_0 - x$,$2x$,$\frac{1}{2}x$
Total pressure $P_t = (P_0 - x) + 2x + \frac{1}{2}x = P_0 + \frac{3}{2}x$
When $50\%$ of the reaction is completed,$x = \frac{P_0}{2}$.
$P_t = P_0 + \frac{3}{2}(\frac{P_0}{2}) = P_0 + \frac{3}{4}P_0 = \frac{7}{4}P_0$.
Thus,the final pressure is $7/4$ times the initial pressure.

(C) The chemical reaction is: $C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2$
At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$ volume.
Number of moles of $CO_2$ produced = $\frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
From the stoichiometry of the reaction,$1 \ mole$ of benzoic acid produces $1 \ mole$ of $CO_2$.
Therefore,moles of benzoic acid = $0.5 \ mol$.
Molar mass of benzoic acid $(C_6H_5COOH)$ = $(6 \times 12) + (6 \times 1) + (2 \times 16) = 72 + 6 + 32 = 122 \ g/mol$.
Mass of benzoic acid $(X)$ = $\text{moles} \times \text{molar mass} = 0.5 \ mol \times 122 \ g/mol = 61 \ g$.

(C) Given that the concentration of $HNO_3$ is $75 \% \ w/w$.
This means $100 \ g$ of the solution contains $75 \ g$ of $HNO_3$.
The density of the solution is $d = 1.25 \ g/mL$.
The volume of $100 \ g$ of the solution is $V = \frac{\text{mass}}{\text{density}} = \frac{100 \ g}{1.25 \ g/mL} = 80 \ mL$.
Thus,$75 \ g$ of $HNO_3$ is present in $80 \ mL$ of the solution.
Therefore,the volume of the solution containing $30 \ g$ of $HNO_3$ is $\frac{80 \ mL}{75 \ g} \times 30 \ g = 32 \ mL$.

(A) The balanced chemical equation is $5I^{-} + IO_3^{-} + 6H^{+} \rightarrow 3I_2 + 3H_2O$.
Moles of $KI$ in $200 \ mL$ of $0.1 \ M$ solution $= 0.1 \times 0.2 = 0.02 \ mol$.
From the stoichiometry,$5 \ mol$ of $I^{-}$ reacts with $1 \ mol$ of $IO_3^{-}$.
So,$0.02 \ mol$ of $I^{-}$ reacts with $0.02 / 5 = 0.004 \ mol$ of $KIO_3$. Statement $(A)$ is correct.
$5 \ mol$ of $I^{-}$ reacts with $6 \ mol$ of $H^{+}$. So,$0.02 \ mol$ of $I^{-}$ reacts with $(6/5) \times 0.02 = 0.024 \ mol$ of $H^{+}$. Statement $(B)$ is incorrect.
$5 \ mol$ of $I^{-}$ produces $3 \ mol$ of $I_2$. So,$0.05 \ mol$ of $I^{-}$ (in $0.5 \ L$) produces $(3/5) \times 0.05 = 0.03 \ mol$ of $I_2$. Statement $(C)$ is incorrect.
In the reaction $IO_3^{-} + 6H^{+} + 5e^{-} \rightarrow \frac{1}{2}I_2 + 3H_2O$,the change in oxidation state of $I$ in $IO_3^{-}$ $(+5)$ to $I_2$ $(0)$ is $5$. Thus,the n-factor is $5$. Equivalent weight $= \frac{\text{Molecular weight}}{5}$. Statement $(D)$ is correct.

(C) The balanced chemical equation is: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$
According to the stoichiometry,$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.
Number of moles of $H_2$ produced = $\frac{220 \ mL}{22400 \ mL \ mol^{-1}} = 0.00982 \ mol$.
Since $1 \ mol$ of $H_2$ requires $1 \ mol$ of $Mg$,moles of $Mg$ required = $0.00982 \ mol$.
Mass of $Mg$ = $\text{moles} \times \text{molar mass} = 0.00982 \ mol \times 24 \ g \ mol^{-1} = 0.2357 \ g$.
Converting to milligrams: $0.2357 \ g \times 1000 = 235.7 \ mg$.

(A) The chemical reaction is: $C_6H_5NO_2 + HNO_3 \xrightarrow{H_2SO_4} C_6H_4(NO_2)_2 + H_2O$
Molar mass of nitrobenzene $(C_6H_5NO_2)$ $= (6 \times 12) + (5 \times 1) + 14 + (2 \times 16) = 123 \ g \ mol^{-1}$
Molar mass of $m-$dinitrobenzene $(C_6H_4N_2O_4)$ $= (6 \times 12) + (4 \times 1) + (2 \times 14) + (4 \times 16) = 168 \ g \ mol^{-1}$
Moles of $m-$dinitrobenzene produced $= \frac{4.2 \ g}{168 \ g \ mol^{-1}} = 0.025 \ mol$
Since $1 \ mol$ of nitrobenzene produces $1 \ mol$ of $m-$dinitrobenzene,moles of nitrobenzene required $= 0.025 \ mol$
Mass of nitrobenzene $(X)$ $= 0.025 \ mol \times 123 \ g \ mol^{-1} = 3.075 \ g$
The nearest integer is $3$.

(C) The concentration of iron required is $12 \ ppm$,which means $12 \ g$ of iron per $10^6 \ g$ of wheat.
Mass of wheat $= 150 \ kg = 150 \times 10^3 \ g$.
Mass of iron required $(w) = \frac{12}{10^6} \times 150 \times 10^3 = 1.8 \ g$.
Molar mass of $FeSO_4 \cdot 7 H_2 O = 56 + 32 + (4 \times 16) + 7 \times (2 \times 1 + 16) = 56 + 32 + 64 + 126 = 278 \ g \ mol^{-1}$.
Let the mass of $FeSO_4 \cdot 7 H_2 O$ be $w_1 \ g$.
Since $1 \ mol$ of $FeSO_4 \cdot 7 H_2 O$ contains $1 \ mol$ of $Fe$,the moles of $Fe$ are equal.
$\frac{w_1}{278} = \frac{1.8}{56}$.
$w_1 = \frac{1.8 \times 278}{56} \approx 8.935 \ g$.
Rounding to the nearest integer,we get $9 \ g$.

(B) The chemical reaction for the decomposition of limestone is:
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
Calculate the mass of pure $CaCO_3$ in $150 \ kg$ of limestone:
$\text{Mass of } CaCO_3 = 150 \ kg \times 0.75 = 112.5 \ kg = 112500 \ g$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g \ mol^{-1}$
Number of moles of $CaCO_3 = \frac{112500 \ g}{100 \ g \ mol^{-1}} = 1125 \ mol$
According to the stoichiometry of the reaction,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CaO$.
Therefore,moles of $CaO = 1125 \ mol$
Molar mass of $CaO = 40 + 16 = 56 \ g \ mol^{-1}$
Mass of $CaO = 1125 \ mol \times 56 \ g \ mol^{-1} = 63000 \ g = 63 \ kg$

(D) According to Dalton's law of partial pressure,the partial pressure of a gas is directly proportional to its mole fraction,which is proportional to the number of moles in a given mass.
$(i)$ The ratio of partial pressure of $N_2$ to $O_2$ is $\frac{P_{N_2}}{P_{O_2}} = \frac{n_{N_2}}{n_{O_2}} = \frac{70/28}{27/32} = \frac{2.5}{0.84375} \approx 2.96$.
$(ii)$ The ratio of partial pressure of $O_2$ to $Ar$ is $\frac{P_{O_2}}{P_{Ar}} = \frac{n_{O_2}}{n_{Ar}} = \frac{27/32}{3/40} = \frac{0.84375}{0.075} = 11.25$.
Thus,the ratios are $2.96$ and $11.25$.

(C) The chemical reaction is: $NaI_{(aq)} + AgNO_{3(aq)} \rightarrow AgI_{(s)} + NaNO_{3(aq)}$
$1$. Calculate the molar mass of $AgI$: $108 + 127 = 235 \ g \ mol^{-1}$.
$2$. Calculate the moles of $AgI$ formed: $n(AgI) = \frac{4.74 \ g}{235 \ g \ mol^{-1}} \approx 0.02017 \ mol$.
$3$. From the stoichiometry,$1 \ mol$ of $NaI$ produces $1 \ mol$ of $AgI$. Therefore,moles of $NaI = 0.02017 \ mol$.
$4$. Calculate the molarity $(M)$: $M = \frac{n}{V(L)} = \frac{0.02017 \ mol}{0.020 \ L} = 1.0085 \ M$.
$5$. The nearest integer value is $1$.

(C) The neutralization reaction is: $2HNO_3 + Ba(OH)_2 \rightarrow Ba(NO_3)_2 + 2H_2O$.
According to the law of equivalence,the number of equivalents of $HNO_3$ must equal the number of equivalents of $Ba(OH)_2$.
$n_{factor} \times M_1 \times V_1 = n_{factor} \times M_2 \times V_2$.
For $HNO_3$,$n_{factor} = 1$. For $Ba(OH)_2$,$n_{factor} = 2$.
$1 \times 0.04 \times V_1 = 2 \times 0.01 \times 50$.
$0.04 \times V_1 = 1$.
$V_1 = \frac{1}{0.04} = 25 \ mL$.

(A) Normality $(N)$ is defined as the number of gram equivalents of solute per liter of solution.
The formula for normality is: $N = \frac{\text{Number of gram equivalents}}{\text{Volume of solution in } L}$
Number of gram equivalents = $\frac{w}{E}$,where $w$ is the mass of solute in grams and $E$ is the equivalent weight.
Substituting this into the normality formula: $N = \frac{w}{E \cdot V(L)}$
Since $V(L) = \frac{V(mL)}{1000}$,we get: $N = \frac{w \cdot 1000}{E \cdot V(mL)}$
Rearranging for $w$: $w = \frac{N \cdot E \cdot V(mL)}{1000}$.

(C) According to the balanced chemical equation: $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$.
From the stoichiometry,$2 \ volumes$ of $NO$ require $1 \ volume$ of $O_2$.
Therefore,$100 \ cm^3$ of $NO$ requires $\frac{1}{2} \times 100 \ cm^3 = 50 \ cm^3$ of pure $O_2$.
Since air contains $20 \%$ $O_2$ by volume,the volume of air required is calculated as: $\text{Volume of air} = \frac{\text{Volume of } O_2}{0.20} = \frac{50 \ cm^3}{0.20} = 250 \ cm^3$.

(A) $A. CaCO_3 \rightarrow CaO + CO_2$. Molar mass of $CaCO_3 = 100 \ g/mol$. $10 \ g = 0.1 \ mol$. $0.1 \ mol \ CO_2$ at $STP = 0.1 \times 22.4 = 2.24 \ L$. So,$A \rightarrow 4$.
$B. Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. Molar mass of $Na_2CO_3 = 106 \ g/mol$. $1.06 \ g = 0.01 \ mol$. $0.01 \ mol \ CO_2$ at $STP = 0.01 \times 22.4 = 0.224 \ L$. So,$B \rightarrow 1$.
$C. C + O_2 \rightarrow CO_2$. Molar mass of $C = 12 \ g/mol$. $2.4 \ g = 0.2 \ mol$. $0.2 \ mol \ CO_2$ at $STP = 0.2 \times 22.4 = 4.48 \ L$. So,$C \rightarrow 2$.
$D. 2CO + O_2 \rightarrow 2CO_2$. Molar mass of $CO = 28 \ g/mol$. $0.56 \ g = 0.02 \ mol$. $0.02 \ mol \ CO$ produces $0.02 \ mol \ CO_2$. $0.02 \ mol \ CO_2$ at $STP = 0.02 \times 22.4 = 0.448 \ L$. So,$D \rightarrow 3$.
Thus,the correct match is $A$ $\rightarrow 4, B$ $\rightarrow 1, C$ $\rightarrow 2, D$ $\rightarrow 3$.

(D) The normality $(N)$ of a solution is given by the formula: $N = \frac{\text{Mass of solute (g)}}{\text{Equivalent weight} \times \text{Volume of solution (L)}}$.
For $H_2SO_4$,the molar mass is $98 \ g/mol$. The basicity of $H_2SO_4$ is $2$,so its equivalent weight is $\frac{98}{2} = 49 \ g/eq$.
Given: $N = 0.05 \ N$,Volume = $1 \ L$,Equivalent weight = $49 \ g/eq$.
Substituting these values: $0.05 = \frac{\text{Mass}}{49 \times 1}$.
Mass = $0.05 \times 49 = 2.45 \ g$.

(D) The combustion reaction for a hydrocarbon $C_xH_y$ is:
$C_xH_y(g) + (x + \frac{y}{4}) O_2(g) \rightarrow x CO_2(g) + \frac{y}{2} H_2O(g)$
Given volumes: $10 \ mL$ of hydrocarbon produces $40 \ mL$ of $CO_2$ and $50 \ mL$ of $H_2O$.
According to Avogadro's Law,volume is proportional to the number of moles at constant temperature and pressure.
Ratio of volumes: $1 : x : \frac{y}{2} = 10 : 40 : 50$
Dividing by $10$: $1 : 4 : 5$
Comparing the coefficients:
$x = 4$
$\frac{y}{2} = 5 \Rightarrow y = 10$
Thus,the hydrocarbon is $C_4H_{10}$.

(A) The decomposition reaction is: $MgCO_{3(s)} \longrightarrow MgO_{(s)} + CO_{2(g)}$
The molar mass of $MgO = 24 + 16 = 40 \ g/mol$.
The number of moles of $MgO$ produced $= \frac{6.0 \ g}{40 \ g/mol} = 0.15 \ mol$.
According to the stoichiometry,$1 \ mol$ of $MgCO_3$ produces $1 \ mol$ of $MgO$.
Therefore,the moles of pure $MgCO_3$ present $= 0.15 \ mol$.
The molar mass of $MgCO_3 = 24 + 12 + (3 \times 16) = 84 \ g/mol$.
The mass of pure $MgCO_3 = 0.15 \ mol \times 84 \ g/mol = 12.6 \ g$.
The percentage purity $= \frac{\text{mass of pure } MgCO_3}{\text{total mass of sample}} \times 100 = \frac{12.6}{20.0} \times 100 = 63 \%$.

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

(C) The chemical reaction for the thermal decomposition of limestone is:
$CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$
From the stoichiometry of the reaction,$100 \ g$ of $CaCO_3$ produces $44 \ g$ of $CO_2$.
Given that the limestone is $80\%$ pure,the mass of pure $CaCO_3$ in $10 \ kg$ of limestone is:
$Mass_{CaCO_3} = 10 \ kg \times 0.80 = 8 \ kg$
Using the molar mass ratio:
$Mass_{CO_2} = \frac{44 \ kg \ CO_2}{100 \ kg \ CaCO_3} \times 8 \ kg \ CaCO_3 = 3.52 \ kg$

(B) The balanced chemical equation is: $CaC_2 + 2 H_2O \rightarrow Ca(OH)_2 + C_2H_2(g)$
The molar mass of $CaC_2 = 40 + (2 \times 12) = 64 \ g/mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $CaC_2$ produces $1 \ mol$ of $C_2H_2$ gas.
At $NTP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$64 \ g$ of $CaC_2$ produces $22.4 \ L$ of $C_2H_2$ gas at $NTP$.
For $100 \ g$ of $CaC_2$,the volume of gas produced is: $\frac{22.4 \ L}{64 \ g} \times 100 \ g = 35 \ L$.

(B) Cane sugar (sucrose) has the chemical formula $C_{12}H_{22}O_{11}$.
When treated with concentrated $H_2SO_4$,it undergoes dehydration.
The chemical reaction is: $C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.
From the stoichiometry of the balanced equation,$1 \text{ mole}$ of sucrose yields $11 \text{ moles}$ of water.

(D) The hydrolysis reaction of sucrose is given by:
$C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 \text{ (glucose)} + C_6H_{12}O_6 \text{ (fructose)}$
From the stoichiometry,$1 \ mol$ of sucrose $(342 \ g)$ produces $1 \ mol$ of glucose $(180 \ g)$.
Therefore,$68.4 \ g$ of sucrose will produce:
$\text{Mass of glucose} = \frac{180 \ g \times 68.4 \ g}{342 \ g} = 36.0 \ g$.

(D) The reaction is: $C_7H_6O_3 (\text{salicylic acid}) + C_4H_6O_3 (\text{acetic anhydride})$ $\rightarrow C_9H_8O_4 (\text{aspirin}) + C_2H_4O_2 (\text{acetic acid})$.
Atom economy is calculated as: $\frac{\text{Molecular mass of desired product}}{\text{Total molecular mass of all reactants}} \times 100$.
Desired product is aspirin $(180 \text{ u})$.
Total mass of reactants = $138 \text{ u} + 102 \text{ u} = 240 \text{ u}$.
Atom economy = $\frac{180}{240} \times 100 = 0.75 \times 100 = 75 \%$.

(B) Atom economy is defined as the ratio of the molecular weight of the desired product to the sum of the molecular weights of all reactants,expressed as a percentage.
Reaction $B$: $C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6$ is an addition reaction where all atoms of the reactants are incorporated into the product.
Atom economy = $\frac{\text{Mol. wt of } C_2H_6}{\text{Mol. wt of } (C_2H_4 + H_2)} \times 100 = \frac{30}{30} \times 100 = 100 \%$.
Since the atom economy is $100 \%$,this reaction exhibits the best atom economy among the given options.

(C) $\text{Percentage atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants}} \times 100$
$\text{Percentage atom economy} = \frac{70 \ u}{140 \ u} \times 100 = 50 \%$

(B) $\text{Percent atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants used in the reaction}} \times 100$
$= \frac{175 \ u}{225 \ u} \times 100$
$= 77.7 \%$

(B) The chemical reaction for the preparation of ethanol from chloroethane is: $C_2H_5Cl + KOH \rightarrow C_2H_5OH + KCl$.
The molar mass of $C_2H_5Cl$ is $64.5 \ g/mol$,$KOH$ is $56 \ g/mol$,$C_2H_5OH$ is $46 \ g/mol$,and $KCl$ is $74.5 \ g/mol$.
The formula for percentage atom economy is: $\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100$.
Total molar mass of reactants = $64.5 + 56 = 120.5 \ g/mol$.
Molar mass of desired product (ethanol) = $46 \ g/mol$.
Atom Economy = $(46 / 120.5) \times 100 \approx 38.17 \%$.
Thus,the correct option is $B$.

(C) The balanced chemical equation for the combustion of benzene is:
$C_6H_6 + \frac{15}{2}O_2 \rightarrow 6CO_2 + 3H_2O$
Molar mass of benzene $(C_6H_6)$ = $(6 \times 12) + (6 \times 1) = 78 \ g/mol$.
Moles of benzene = $\frac{15.6 \ g}{78 \ g/mol} = 0.2 \ mol$.
From the stoichiometry,$1 \ mol$ of $C_6H_6$ requires $\frac{15}{2} = 7.5 \ mol$ of $O_2$.
Therefore,$0.2 \ mol$ of $C_6H_6$ requires $0.2 \times 7.5 = 1.5 \ mol$ of $O_2$.
Mass of $O_2$ required = $1.5 \ mol \times 32 \ g/mol = 48 \ g$.

(B) The combustion reaction of carbon is: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
From the stoichiometry,$1 \ mol$ of $C$ produces $1 \ mol$ of $CO_{2(g)}$.
Molar mass of $C = 12 \ g/mol$.
Number of moles of $C = \frac{6 \ g}{12 \ g/mol} = 0.5 \ mol$.
Since $1 \ mol$ of $C$ gives $1 \ mol$ of $CO_{2(g)}$,$0.5 \ mol$ of $C$ will produce $0.5 \ mol$ of $CO_{2(g)}$.
At $STP$,the molar volume of an ideal gas is $22.4 \ dm^3/mol$.
Volume of $CO_{2(g)} = 0.5 \ mol \times 22.4 \ dm^3/mol = 11.2 \ dm^3$.

(B) The balanced chemical equation for the thermal decomposition of potassium chlorate is: $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$
From the stoichiometry,$2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
$2 \text{ moles}$ of $KClO_3 = 2 \times 122.5 \ g = 245 \ g$.
$3 \text{ moles}$ of $O_2$ at $STP$ occupy $3 \times 22.4 \ dm^3 = 67.2 \ dm^3$.
Thus,$245 \ g$ of $KClO_3$ produces $67.2 \ dm^3$ of $O_2$.
To find the mass $(x)$ required to produce $22.4 \ dm^3$ of $O_2$:
$x = \frac{245 \ g \times 22.4 \ dm^3}{67.2 \ dm^3} = 81.67 \ g$.

(B) The number of moles of a gas $(n)$ is given by the formula: $n = \frac{\text{Volume of gas at STP}}{22.4 \ dm^3 \ mol^{-1}}$.
From the given volume,$n(H_2) = \frac{4.48 \ dm^3}{22.4 \ dm^3 \ mol^{-1}} = 0.2 \ mol$.
According to the balanced chemical equation: $Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2_{(aq)}} + H_{2_{(g)}} \uparrow$.
$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.
Therefore,the moles of $Mg$ required to produce $0.2 \ mol$ of $H_2$ gas is $0.2 \ mol$.
Mass of $Mg = \text{moles} \times \text{molar mass} = 0.2 \ mol \times 24 \ g \ mol^{-1} = 4.8 \ g$.

(B) For acid-base neutralization,the number of gram-equivalents of acid must equal the number of gram-equivalents of base.
The formula for gram-equivalents is: $\text{Gram-equivalents} = \text{Normality} (N) \times \text{Volume} (V \text{ in } L)$.
Given:
Normality of $HCl$ $(N)$ = $0.1 \ N$ (decinormal).
Volume of $HCl$ $(V)$ = $25 \ cm^{3} = 25 \times 10^{-3} \ L = 0.025 \ L$.
Therefore,gram-equivalents of $HCl = 0.1 \times 0.025 = 0.0025$.
Since the reaction is $HCl + NaOH \rightarrow NaCl + H_2O$,the gram-equivalents of $NaOH$ required = gram-equivalents of $HCl = 0.0025$.

(C) Given: Initial volume $V_{1} = 2.5 \ cm^{3} = 2.5 \times 10^{-3} \ dm^{3}$.
Initial molarity $M_{1} = 0.2 \ M$.
Since $H_{2}SO_{4}$ is a dibasic acid,its initial normality $N_{1} = M_{1} \times \text{basicity} = 0.2 \times 2 = 0.4 \ N$.
Final volume $V_{2} = 0.5 \ dm^{3}$.
Using the dilution equation $N_{1}V_{1} = N_{2}V_{2}$:
$0.4 \ N \times 2.5 \times 10^{-3} \ dm^{3} = N_{2} \times 0.5 \ dm^{3}$.
$N_{2} = \frac{0.4 \times 2.5 \times 10^{-3}}{0.5} = \frac{1 \times 10^{-3}}{0.5} = 2 \times 10^{-3} \ N = 0.002 \ N$.

(B) The balanced chemical equation for the reaction is:
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$
According to Avogadro's law,at constant temperature and pressure,the volume ratio of gaseous reactants and products is equal to their stoichiometric mole ratio.
From the equation,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.
Given $10 \ dm^3$ of $N_2$ and $30 \ dm^3$ of $H_2$,the stoichiometric ratio is exactly $1:3$.
Therefore,the volume of $NH_3$ formed is $2 \times 10 \ dm^3 = 20 \ dm^3$.

(B) At $STP$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \ dm^3 \ mol^{-1}$.
Given the amount of substance $n = 0.5 \ mol$.
The volume $V$ is calculated as $V = n \times \text{molar volume}$.
$V = 0.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 11.2 \ dm^3$.
Therefore,the correct option is $B$.

(C) The balanced chemical equation for the thermal decomposition of potassium chlorate is:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
From the stoichiometry,$2 \ mol$ of $KClO_3$ produces $3 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$.
Therefore,$3 \ mol$ of $O_2$ occupies $3 \times 22.4 \ dm^3 = 67.2 \ dm^3$.
Mass of $2 \ mol$ of $KClO_3 = 2 \times 122.5 \ g = 245 \ g$.
Since $67.2 \ dm^3$ of $O_2$ is produced by $245 \ g$ of $KClO_3$,
$5.6 \ dm^3$ of $O_2$ will be produced by:
$\frac{245 \ g}{67.2 \ dm^3} \times 5.6 \ dm^3 = 20.416 \ g \approx 20.40 \ g$.

(D) The molar mass of anhydrous calcium chloride $(CaCl_2)$ is $40 + 2 \times 35.5 = 111 \ g/mol$.
Number of moles of $CaCl_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{222 \ g}{111 \ g/mol} = 2 \ mol$.
Each mole of $CaCl_2$ dissociates to give $2 \ mol$ of $Cl^{-}$ ions.
Total moles of $Cl^{-}$ ions $= 2 \times 2 \ mol = 4 \ mol$.
Number of $Cl^{-}$ ions $= \text{moles} \times N_A = 4 \ N_A$.

(C) The molar mass of dinitrogen $(N_2)$ is $2 \times 14 = 28 \ g/mol$.
Number of moles $(n)$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{56 \ g}{28 \ g/mol} = 2 \ mol$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,the volume of $2 \ mol$ of $N_2$ = $2 \times 22.4 \ L = 44.8 \ L$.

(A) The chemical equation for the combustion of carbon is: $C + O_2 \rightarrow CO_2$
From the stoichiometry of the reaction,$12 \ g$ of $C$ produces $44 \ g$ of $CO_2$.
Therefore,$0.6 \ g$ of $C$ will produce: $\frac{44 \times 0.6}{12} = 2.2 \ g$ of $CO_2$.
The molar mass of $CO_2$ is $44 \ g/mol$.
Number of moles of $CO_2 = \frac{2.2 \ g}{44 \ g/mol} = 0.05 \ mol$.
Number of molecules of $CO_2 = \text{moles} \times N_A = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22}$ molecules.

(B) The dehydration reaction of sugar $(C_{12}H_{22}O_{11})$ by concentrated sulphuric acid is given by:
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \ g/mol$.
Number of moles of sugar = $\frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
From the stoichiometry of the reaction,$1 \ mol$ of sugar produces $12 \ mol$ of carbon (sugar charcoal).
Therefore,$0.1 \ mol$ of sugar produces $0.1 \times 12 = 1.2 \ mol$ of carbon.
Mass of charcoal = $1.2 \ mol \times 12 \ g/mol = 14.4 \ g$.