What volume of hydrogen gas measured at s.t.p. is evolved when 13.0\text{ g} of pure zinc … — JAMB UTME

What volume of hydrogen gas measured at s.t.p. is evolved when

13.0\text{ g}

of pure zinc completely reacts with excess dilute hydrochloric acid according to the equation below?

\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)

[\text{Zn} = 65, \text{Molar gas volume at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

4.48\text{ dm}^3

The balanced chemical equation indicates a 1:1 molar relationship between zinc and hydrogen gas. Reacting

13.0\text{ g}

of zinc corresponds to

\frac{13.0}{65} = 0.20\text{ mol}

of zinc, yielding

0.20\text{ mol}

of hydrogen gas. Multiplying by the molar volume at s.t.p. (

22.4\text{ dm}^3\text{ mol}^{-1}

) gives

0.20 \times 22.4 = 4.48\text{ dm}^3

Calculate the number of moles of zinc reacted

\text{Moles of Zn} = \frac{13.0\text{ g}}{65\text{ g mol}^{-1}} = 0.20\text{ mol}

Converting given mass of reactant to moles using relative atomic mass.

Determine the moles of hydrogen gas produced from the balanced equation

\text{Mole ratio of Zn to H}_2 = 1 : 1

, so

\text{moles of H}_2 = 0.20\text{ mol}

Applying mole ratio constraints from the chemical equation.

Calculate the volume of hydrogen gas at s.t.p.

\text{Volume of H}_2 = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

Multiplying moles of gas by standard molar volume at s.t.p.

Mass-Volume stoichiometric calculation at standard temperature and pressure (s.t.p.)

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