Question

1 gm of an alloy of Al and Mg reacts with excess H to form AlCl${}_3$, MgCl${}_2$ and H${}_2$. The evolved H${}_2$ collected over mercury at $0{}^{o}C$ occupied 1200 mL at 699 mm Hg. What is the composition of alloy?

Answer 1

The ionic reaction equation for the reaction is :

1.) Mg(s) + 2H⁺ (aq) $\to$ Mg²⁺(aq) + H₂(g)

Mole ratio is 1 : 2

2.) 2Al(s) + 6H⁺(aq) $\to$ 2Al³⁺(aq) + 3H₂(g)

Mole ratio is 1 : 3

1 mole of H₂ = 22.4 litres at stp

1.12 / 22.4 = 0.05 moles.

Moles of Mg = 0.05 / 2 = 0.025 moles

Mass is :

0.025 × 24 = 0.6g

Moles of Al = 0.05 / 3 = 0.01667 moles.

Mass is :

0.01667 × 27 = 0.45 g

Composition of the alloy:

0.45 + 0.6 = 1.05g