The volume of hydrogen that is formed by the interaction of 9 g of technical aluminum (15% of impurities) with an excess of sodium hydroxide.
Data: m is the mass of technical aluminum (m = 9 g); ωпр – mass fraction of impurities in the taken technical aluminum (ωпр = 15% = 0.15).
Const: Vm – molar volume (at normal conditions Vm = 22.4 l / mol); МAl – molar mass of aluminum (МAl ≈ 27 g / mol).
1) The mass of pure aluminum: mAl = m * (1 – ωpr) = 9 * (1 – 0.15) = 7.65 g.
2) Amount of aluminum substance: νAl = mAl / MAl = 7.65 / 27 ≈ 0.283 (3) mol.
3) The reaction level: 2Аl (aluminum) + 2NaOH (sodium hydroxide) + 2Н2О (water) = 2NaAlO2 (sodium aluminate) + 3Н2 ↑ (hydrogen).
4) Amount of the substance of the formed hydrogen: νH2 / νAl = 3/2 and νH2 = 3 * νAl / 2 = 3 * 0.283 (3) / 2 = 0.425 mol.
5) The volume of the generated hydrogen: VH2 = νH2 * Vm = 0.425 * 22.4 = 9.52 liters.
Answer: 9.52 liters of hydrogen formed.
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