When 9 g of an aluminum-magnesium alloy was dissolved in an alkali solution, 9.6 liters

| January 27, 2021 | education

When 9 g of an aluminum-magnesium alloy was dissolved in an alkali solution, 9.6 liters of hydrogen were released. Determine the mass fraction of aluminum in the alloy.

1. Since we are not indicated which alkali entered into the reaction, take, for example, NaOH and compose the reaction equations:

a) Mg + NaOH = X (no reaction).

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

2. Find the amount of hydrogen (Vm is a constant equal to 22.4 mol / l):

n (H2) = V (H2) / Vm (H2) = 9.6 L / 22.4 mol / L = 0.429 mol.

3. Using the equation, compose the ratio of the amounts of hydrogen and aluminum:

n (Al) / 2 = n (H2) / 3.

n (Al) = 2 * n (H2) / 3 = 2 * 0.429 mol / 3 = 0.286 mol.

4. Find the mass and mass fraction of aluminum:

m (Al) = n (Al) * M (Al) = 0.286 mol * 27 g / mol = 7.72 g.

ω (Al) = (m (Al) / m (solution)) * 100% = (7.72 g / 9 g) * 100% = 85.78%.

ω (Mg) = 100% – ω (Al) = 100% – 85.78% = 14.22%.

Answer: ω (Al) = 85.78%; ω (Mg) = 14.22%.



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