When processing a mixture of iron sawdust and aluminum with an excess of 10% hydrochloric acid solution

When processing a mixture of iron sawdust and aluminum with an excess of 10% hydrochloric acid solution, 8.96 liters of gas were released. When processing the same sample of the mixture with an excess of 10% sodium hydroxide solution, 6.72 liters of gas were released. Determine the mass fraction of iron in the original mixture of metals.

Fe + 2HCl = FeCl2 + H2.

2Al + 6HCl = 2AlCl3 + 3H2.

2Al + NaOH + 6H2O = 2Na [Al (OH) 4] + 3H2.

Let’s find the amount of substance H2.

n = V: Vn.

n = 6.72 L: 22.4 L / mol = 0.3 mol.

According to the reaction equation, there is 2 mol of Al for 3 mol of H2. Substances are in quantitative ratios 3: 2 = 1.5: 1.

The amount of Al substance is 1.5 times less than the amount of H2 substance.

n (Al) = 2/3 n (H2) = 0.3: 1.5 = 0.2 mol.

Let us find the mass of Al.

M (Al) = 27 g / mol.

m = n × M.

m = 27 g / mol × 0.2 mol = 5.4 g.

V (H2) = 8.96 – 6.72 = 2.24 liters.

n (H2) = 2.24 L: 22.4 L / mol = 0.1 mol.

n (Fe) = n (H2) = 0.1 mol.

Let us find the mass of Fe.

M (Fe) = 56 g / mol.

m = n × M.

m = 56 g / mol × 0.1 mol = 5.6 g.

m (mixture) = m (Fe) + m (Al) = 5.6 g + 5.4 g = 11 g.

w (Fe) = (5.6g: 11g) × 100% = 50.91%.

Answer: 50.91%.