What volume of hydrogen gas, measured at STP, is produced when 4.8\text{ g} of magnesium r… — JAMB UTME

What volume of hydrogen gas, measured at STP, is produced when $4.8\text{ g}$ of magnesium ribbon reacts completely with excess dilute tetraoxosulfate(VI) acid according to the chemical equation below?

Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

$[Mg = 24, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

Answer: 4.48 dm^3

Answer

The volume of hydrogen gas produced at STP is

4.48\text{ dm}^3

From the stoichiometric relationship in the balanced reaction

Mg(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2(g)

1\text{ mol}

Mg

(

24\text{ g}

) produces

1\text{ mol}

H_2

gas (

22.4\text{ dm}^3

at STP). For

4.8\text{ g}

Mg

, the number of moles is

\frac{4.8}{24} = 0.20\text{ mol}

. Multiplying by the molar gas volume gives

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

H_2

gas.

Step-by-Step Solution

Calculate the amount in moles of magnesium (

Mg

) reacted.

n(Mg) = \frac{4.8\text{ g}}{24\text{ g mol}^{-1}} = 0.20\text{ mol}

Dividing given mass by relative atomic mass yields the number of moles.

Determine the amount in moles of hydrogen gas (

H_2

) produced.

n(H_2) = 0.20\text{ mol}

The balanced chemical equation shows a

1:1

stoichiometric molar ratio between

Mg

and

H_2

Calculate the volume of

H_2

gas at STP.

V(H_2) = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

At standard temperature and pressure (STP), one mole of any gas occupies

22.4\text{ dm}^3

Key Concept

Mass-Volume stoichiometric calculation at STP

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic