Aluminum carbide was dissolved in 150 g of a 12% hydrochloric acid solution.

| February 6, 2021 | education

Aluminum carbide was dissolved in 150 g of a 12% hydrochloric acid solution. The methane released during this took up a volume of 2.24 liters. Calculate the mass fraction of salt in the resulting solution

Moles of methane:
n (CH4) = V / Vm = 2.24 / 22.4 = 0.1 mol.
Acid weight without impurities:
m (pure HCl) = m * W = 150 * 12% = 18 g.
Number of moles of acid:
n (HCl) = m / Mr = 18 / 36.5 = 0.5 mol.
The amount of acid reacted will be x:
x / 12 = 0.1 / 3.
x = 0.4 mol HCl.
Remaining acid:
n1 (HCl) = 0.5 – 0.4 = 0.1 mol.
Weight of unreacted acid:
m1 (HCl) = n1 * Mr = 0.1 * 36.5 = 3.65 g.
Mass of unreacted water:
m (H2O) = m (solution) -m (HCl) = 150-18 = 132 g.
Moles of salt:
n (AlCl3) = n (CH4) * 4/3 = 4/30 mol.
Salt mass:
m (AlCl3) = n (AlCl3) * M (AlCl3) = 4/30 * 133.5 = 17.8 g.
w (mass fraction of AlCl3) = m (AlCl3) / m (solution) = 17.8 / (3.65 + 132 + 17.8) = 11.6%.



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