After treatment of 118.5 g of a mixture of aluminum and calcined (chemically passive) alumina with an excess of concentrated sodium

After treatment of 118.5 g of a mixture of aluminum and calcined (chemically passive) alumina with an excess of concentrated sodium hydroxide solution, 8.4 liters of gas were collected. Determine the mass fraction (%) of aluminum oxide in the initial mixture.

Al2O3
2Al + 2NaOH + 6H2O → 2Na [Al (OH) 4] + 3H2 ↑
n (H2) = 8.4 / 22.4 = 0.375 mol
n (Al) = 0.375 / 3 x 2 = 0.25 mol
m (Al) = 0.25 x 27 = 6.75 g
m (Al) = 0.25 x 27 = 6.75 g
m (Al2O3) = 118.5 – 6.75 = 111.75 g
ω (Al2O3) = 111.75 / 118.5 x 100% = 94.3%
Answer: the mass fraction of aluminum in the initial mixture is 94.3%.

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