The alloy contains 85% magnesium and 15% aluminum. A weighed portion of 8.52 g of the alloy was treated

| December 16, 2020 | education

The alloy contains 85% magnesium and 15% aluminum. A weighed portion of 8.52 g of the alloy was treated with hydrochloric acid. Calculate the amount of hydrogen evolved under normal conditions.

2Al + 6HCl = 2AlCl3 + 3H2
Mg + 2HCl = MgCl2 + H2
m (Al) = m sample * w (Al) = 8.52 * 0.15 = 1.278 g
m (Mg) = m sample * w (Mg) = 8.52 * 0.85 = 7.242 g
n (Al) = m / M = 1.278 / 27 = 0.0473 mol
n (Al): n1 (H2) = 2: 3 (see numbers before the formulas of substances in the equation)
n1 (H2) = 0.0437 / 2 * 3 = 0.071 mol
n (Mg) = m / M = 7.242 / 24 = 0.3 mol
n (Mg): n2 (H2) = 1: 1 (see numbers before the formulas of substances in the equation)
n2 (H2) = 0.3 mol
n (H2) = n1 (H2) + n2 (H2) = 0.071 + 0.3 = 0.371 mol
V (H2) = n * Vm = 0.371 * 22.4 = 8.31 l
Answer: 8.31 L



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