a) A trivalent metal oxide contain 31.6\% of oxygen. What is the atomic mass of the metal? [2] b) 16.8 gm of NaHCO 3 , when heated produces 9gm of Na 2CO 3 according to the following reaction:…

a) Atomic mass of the metal

Let $M$ be the atomic mass of the metal. The metal oxide is trivalent, so its formula is $\mathrm{M}{}_2\mathrm{O}{}_3$.

Given that the oxide contains 31.6% oxygen by mass, the percentage of the metal is: $$100\%-31.6\%=68.4\%$$

The molar mass of $\mathrm{M}{}_2\mathrm{O}{}_3$ is: $$2M+3\times 16=2M+48$$

Since the metal constitutes 68.4% of the mass: $$\frac{2M}{2M+48}=\frac{68.4}{100}$$ $$200M=68.4(2M+48)$$ $$200M=136.8M+3283.2$$ $$63.2M=3283.2$$ $$M=\frac{3283.2}{63.2}=52$$

The atomic mass of the metal is 52.

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b) $\mathrm{NaHCO}{}_3$ decomposition

i) Percentage purity of $\mathrm{NaHCO}{}_3$

Given:

• Mass of impure $\mathrm{NaHCO}{}_3=16.8\text{ g}$
• Mass of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ produced = 9 g

The balanced reaction is: $$2\mathrm{NaHCO}{}_3\to \mathrm{Na}{}_2\mathrm{CO}{}_3+\mathrm{H}{}_2\mathrm{O}+\mathrm{CO}{}_2$$

Molar masses:

• $\mathrm{NaHCO}{}_3=23+1+12+3\times 16=84\text{ g/mol}$
• $\mathrm{Na}{}_2\mathrm{CO}{}_3=2\times 23+12+3\times 16=106\text{ g/mol}$

From the reaction, 2 moles of $\mathrm{NaHCO}{}_3$ produce 1 mole of $\mathrm{Na}{}_2\mathrm{CO}{}_3$. Thus, the theoretical mass of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ from pure $\mathrm{NaHCO}{}_3$ is: $$\text{Mass of }\mathrm{Na}{}_2\mathrm{CO}{}_3=\frac{106}{2\times 84}\times \text{mass of pure }\mathrm{NaHCO}{}_3$$

Let $x$ be the mass of pure $\mathrm{NaHCO}{}_3$: $$9=\frac{106}{168}\times x$$ $$x=\frac{9\times 168}{106}=14.72\text{ g}$$

Percentage purity: $$\frac{14.72}{16.8}\times 100=87.6\%$$

The percentage purity of $\mathrm{NaHCO}{}_3$ is 87.6%.

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ii) Volume of $\mathrm{CO}{}_2$ produced

From the reaction, 2 moles of $\mathrm{NaHCO}{}_3$ produce 1 mole of $\mathrm{CO}{}_2$.

Mass of pure $\mathrm{NaHCO}{}_3$ = 14.72 g Moles of $\mathrm{NaHCO}{}_3$: $$n=\frac{14.72}{84}=0.175\text{ mol}$$

Moles of $\mathrm{CO}{}_2$ produced: $$n_{\mathrm{CO}{}_2}=\frac{0.175}{2}=0.0875\text{ mol}$$

Using the ideal gas equation $PV=nRT$: $$V=\frac{nRT}{P}$$ Given:

• $P=1\text{ atm}$
• $T=27^{\circ }\text{C}=300\text{ K}$
• $R=0.0821{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$

$$V=\frac{0.0875\times 0.0821\times 300}{1}=2.14\text{ L}$$

The volume of $\mathrm{CO}{}_2$ produced is 2.14 L.