Question

Predict the mass of hydrogen gas produced when 2.73 g of aluminum reacts in a single replacement reaction with excess sulphuric acid.

2Al + 3H sub 2 SO sub 4, reaction arrow, 3H sub 2 + Al sub 2 (SO sub 4 ) sub 3.

Answer 1

Answer on Question #52447 - Chemistry – Inorganic Chemistry

Question

Predict the mass of hydrogen gas produced when 2.73 g of aluminum reacts in a single replacement reaction with excess sulphuric acid.

$$2 \mathrm{Al} + 3 \mathrm{H_2SO_4} \rightarrow 3 \mathrm{H_2} + \mathrm{Al_2(SO_4)_3}$$

Answer:

Molar mass of Al is 27.3 g/mol. Molar mass of H₂ is 2 g/mol.

Number of moles of aluminum is:

$$n(Al) = \frac{m(Al)}{M(Al)} = \frac{2.73}{27.3} = 0.1\ \text{mol}$$

Make a proportion:

- 2 mol of Al produce 3 mol of H₂

- 0.1 mol of Al – x g of H₂

- $x = \frac{0.1 \cdot 3}{2} = 0.15\ \text{mol}$

The mass of hydrogen gas produced is:

$$m(H_2) = n(H_2) \cdot M(H_2) = 0.15 \cdot 2 = 0.3\ \text{g}$$

Answer: $m(H_2) = 0.3\ \text{g}$

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