What volume of hydrogen gas, measured at STP, is liberated when 5.4\text{ g} of aluminium … — JAMB UTME

What volume of hydrogen gas, measured at STP, is liberated when

5.4\text{ g}

of aluminium react completely with excess hydrochloric acid according to the following equation?

2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2(g)

[Relative atomic mass:

\text{Al} = 27

; Molar volume of gas at STP

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

]

The volume of hydrogen gas liberated at STP is

6.72\text{ dm}^3

The option specifying

6.72\text{ dm}^3

is correct because

5.4\text{ g}

of aluminium corresponds to

0.20\text{ mol}

. Based on the stoichiometric ratio

2\text{Al}:3\text{H}_2

0.20\text{ mol}

of aluminium liberates

0.30\text{ mol}

of hydrogen gas. Multiplying

0.30\text{ mol}

by the molar gas volume at STP (

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

) yields exactly

6.72\text{ dm}^3

Calculate the amount in moles of aluminium that reacted

\text{Moles of Al} = \frac{5.4\text{ g}}{27\text{ g mol}^{-1}} = 0.20\text{ mol}

Converting mass of reactant to moles is required to apply the stoichiometric mole ratio.

Determine the moles of hydrogen gas produced using the balanced chemical equation

\text{Moles of H}_2 = 0.20\text{ mol Al} \times \frac{3\text{ mol H}_2}{2\text{ mol Al}} = 0.30\text{ mol H}_2

The balanced chemical equation shows that

2\text{ mol}

\text{Al}

produce

3\text{ mol}

\text{H}_2

Calculate the volume of hydrogen gas produced at STP

\text{Volume of H}_2 = 0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

Mass-Volume Stoichiometric Calculations at STP

Question