Stoichiometry

Moles of CO$_2$ required:

$n = \frac{0.220\ \text{g}}{44\ \text{g mol}^{-1}} = 5.0\times10^{-3}\ \text{mol}$

From $\mathrm{C + O_2 \rightarrow CO_2}$, 1 mol O$_2$ gives 1 mol CO$_2$, so O$_2$ needed $= 5.0\times10^{-3}$ mol.

$\text{Molecules} = 5.0\times10^{-3}\times 6.022\times10^{23} = 3.01\times10^{21}$

About $3.01\times10^{21}$ oxygen molecules are required.

Moles of glucose ($\mathrm{C_6H_{12}O_6}$, $M=180$): $\dfrac{45}{180}=0.25$ mol Moles of water ($M=18$): $\dfrac{36}{18}=2$ mol

Hydrogen (atoms): from glucose: $0.25\times12=3$ mol; from water: $2\times2=4$ mol; total $=7$ mol $7\times6.022\times10^{23}=4.2\times10^{24}\ \text{H atoms}$

Oxygen (atoms): from glucose: $0.25\times6=1.5$ mol; from water: $2\times1=2$ mol; total $=3.5$ mol $3.5\times6.022\times10^{23}=2.1\times10^{24}\ \text{O atoms}$

(Strictly these are atoms, as no free H₂ or O₂ molecules exist in the solution; counted as H₂ and O₂ units they would be half these numbers, i.e. $2.1\times10^{24}$ and $1.05\times10^{24}$.)

$\mathrm{2Mg + O_2 \rightarrow 2MgO}$

Moles of Mg $=\dfrac{2}{24}=0.0833$; moles of O₂ $=\dfrac{3}{32}=0.0938$. Oxygen needed for 0.0833 mol Mg $=0.0833/2=0.0417$ mol, which is less than what is present.

(i) Magnesium is the limiting reactant.

(ii) Oxygen left over $=0.0938-0.0417=0.0521$ mol ($\approx1.67$ g).

(iii) Mol of MgO $=0.0833$ (same as Mg), so mass $=0.0833\times40=3.33$ g.

(iv) $\mathrm{MgO + H_2SO_4 \rightarrow MgSO_4 + H_2O}$ (1 : 1), so moles of H₂SO₄ $=0.0833$: $0.0833\times98=8.17\ \mathrm{g}$

(a) A trivalent metal $\mathrm{M}$ forms the oxide $\mathrm{M_2O_3}$. Oxygen is $31.6\%$ of the oxide, and $3\ \mathrm{O} = 3\times16 = 48$.

$\text{Molar mass of oxide} = \frac{48}{31.6}\times100 = 151.9\ \mathrm{g\,mol^{-1}}$

$2M = 151.9 - 48 = 103.9 \Rightarrow M = 51.95 \approx 52$

The atomic mass of the metal is about 52 (the metal is chromium).

(b) $\mathrm{2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2}$

Molar masses: $\mathrm{NaHCO_3} = 84$, $\mathrm{Na_2CO_3} = 106$.

(i) Percentage purity

$2\times84 = 168$ g of $\mathrm{NaHCO_3}$ gives $106$ g of $\mathrm{Na_2CO_3}$. So the $\mathrm{NaHCO_3}$ needed for $9$ g of $\mathrm{Na_2CO_3}$ is

$\frac{168}{106}\times9 = 14.26\ \text{g}$

$\text{Purity} = \frac{14.26}{16.8}\times100 = \mathbf{84.9\%}$

(ii) Volume of $\mathrm{CO_2}$

$n(\mathrm{CO_2}) = n(\mathrm{Na_2CO_3}) = \frac{9}{106} = 0.0849\ \text{mol}$

At $P = 1$ atm and $T = 273+27 = 300$ K:

$V = \frac{nRT}{P} = \frac{0.0849\times0.0821\times300}{1} = \mathbf{2.09\ L}$

(d) 6

One molecule of water (H₂O) contains 2 hydrogen atoms + 1 oxygen atom = 3 atoms. So two molecules contain $2 \times 3 = 6$ atoms.

(b) 1

The reaction is $\mathrm{N_{2} + 3H_{2} \rightarrow 2NH_{3}}$, so 1 mole of nitrogen gives 2 moles of ammonia. Hence 0.5 mole of nitrogen gives $0.5 \times 2 = 1$ mole of ammonia.

Given: $I = 0.1\ \text{A}$, $t = 30\ \text{min} = 1800\ \text{s}$, efficiency $= 75\%$, Na$^+ + e^- \rightarrow$ Na (so $n = 1$, $M = 23\ \text{g mol}^{-1}$).

Charge passed:

$Q = It = 0.1 \times 1800 = 180\ \text{C}$

Theoretical mass (Faraday's law, $F = 96500\ \text{C mol}^{-1}$):

$m = \frac{Q}{F}\times\frac{M}{n} = \frac{180}{96500}\times 23 = 0.0429\ \text{g}$

Actual mass at 75% efficiency:

$m_{\text{actual}} = 0.0429 \times \frac{75}{100} = 0.0322\ \text{g}$

Mass of sodium deposited $\approx 0.032$ g.

Oxygen is the limiting reagent because 2 mol NH₃ needs $\frac{5}{4}\times 2 = 2.5$ mol O₂, but only 2 mol O₂ is available.

From the equation: $5\text{ mol O}_2 \rightarrow 4\text{ mol NO}$ $2\text{ mol O}_2 \rightarrow \frac{4}{5}\times 2 = 1.6\text{ mol NO}$

Molar mass of NO $=14+16=30\text{ g/mol}$

$\text{Mass of NO} = 1.6 \times 30 = 48\text{ g}$

Mass of nitric oxide produced = 48 g

It determines the maximum amount of product formed.

The limiting reactant is the reactant that is completely consumed first in a reaction. Since the reaction stops once this reactant runs out, the amount of limiting reactant decides the theoretical yield of the product; the other reactant (excess reactant) is left partly unreacted.

The reaction is $4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$ (with oxygen in excess, as the question intends).

From the equation, 4 mol NH$_3$ give 6 mol H$_2$O, so 2 mol NH$_3$ give

$\frac{6}{4}\times 2 = 3\ \text{mol H}_2\text{O}$

Theoretical mass of water $= 3 \times 18 = 54\ \text{g}$.

$\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}}\times 100 = \frac{50}{54}\times 100 = 92.6\%$

Percentage yield $\approx 92.6\%$.

Balanced equation:

$\text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s)$

1 mol NH$_3$ reacts with 1 mol HCl, so 2 mol NH$_3$ need 2 mol HCl.

Given: $n = 2$ mol, $T = 20 + 273 = 293$ K, $P = \dfrac{750}{760}\ \text{atm} = 0.9868$ atm, $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$.

$PV = nRT \;\Rightarrow\; V = \frac{nRT}{P}$

$V = \frac{2 \times 0.0821 \times 293}{0.9868} = \frac{48.11}{0.9868} \approx 48.75\ \text{L}$

Volume of HCl gas required $\approx 48.75$ L.