A 5.00\text{ g} sample of impure zinc granules reacts completely with excess dilute tetrao… — JAMB UTME

A $5.00\text{ g}$ sample of impure zinc granules reacts completely with excess dilute tetraoxosulfate(VI) acid to produce $1.12\text{ dm}^3$ of hydrogen gas at s.t.p. What is the percentage purity of the zinc sample? $[\text{Zn} = 65, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]$

A
$30.0\%
B
$60.7\%
$65.0\%Answer
D
$35.0\%

Answer

65.0\%

The balanced chemical equation

\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2

shows a

1:1

mole ratio between zinc and hydrogen gas. At s.t.p.,

1.12\text{ dm}^3

\text{H}_2

represents

\frac{1.12}{22.4} = 0.05\text{ mol}

. Thus,

0.05\text{ mol}

of pure zinc was present, corresponding to

0.05 \times 65 = 3.25\text{ g}

. Dividing the pure zinc mass by the total sample mass (

5.00\text{ g}

) gives

\frac{3.25}{5.00} \times 100\% = 65.0\%

Step-by-Step Solution

Calculate the amount of hydrogen gas produced in moles at s.t.p.

\text{Moles of H}_2 = \frac{1.12\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.05\text{ mol}

The reaction produces

1\text{ mol}

\text{H}_2

for every

1\text{ mol}

of pure zinc reacted:

\text{Zn (s)} + \text{H}_2\text{SO}_4\text{ (aq)} \rightarrow \text{ZnSO}_4\text{ (aq)} + \text{H}_2\text{ (g)}

Determine the mass of pure zinc that reacted.

\text{Mass of pure Zn} = 0.05\text{ mol} \times 65\text{ g mol}^{-1} = 3.25\text{ g}

Molar mass converts the stoichiometric mole amount to pure mass of the element.

Calculate the percentage purity of the zinc sample.

\text{Percentage Purity} = \left( \frac{3.25\text{ g}}{5.00\text{ g}} \right) \times 100\% = 65.0\%

Percentage purity is the ratio of pure substance mass to total sample mass expressed as a percentage.

Key Concept

Calculating percentage purity from stoichiometric gas yields at s.t.p.

Estimated Time:1m 30s

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