# Set the volume of gas collected after adding 0.3 mol of aluminum to a 20% potassium

Set the volume of gas collected after adding 0.3 mol of aluminum to a 20% potassium hydroxide solution (volume 160 ml, density 1.19 g / ml.).

Given:
n (Al) = 0.3 mol
ω (KOH) = 20%
V solution (KOH) = 160 ml
ρ solution (KOH) = 1.19 g / ml

Find:
V (gas) -?

Solution:
1) 2Al + 2KOH + 6H2O => 2K [Al (OH) 4] + 3H2 ↑;
2) m solution (KOH) = ρ solution (KOH) * V solution (KOH) = 1.19 * 160 = 190.4 g;
3) m (KOH) = ω (KOH) * m solution (KOH) / 100% = 20% * 190.4 / 100% = 38.08 g;
4) n (KOH) = m (KOH) / M (KOH) = 38.08 / 56 = 0.68 mol;
5) n (H2) = n (Al) * 3/2 = 0.3 * 3/2 = 0.45 mol;
6) V (H2) = n (H2) * Vm = 0.45 * 22.4 = 10.08 l.

Answer: The volume of H2 is 10.08 liters.

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