What volume of hydrogen will be released during the interaction of 2.7 g of 25% hydrochloric acid with the amount of aluminum substance required for the reaction? How much is this substance?
m solution (HCl) = 2.7 g
ω (HCl) = 25%
V (H2) -?
n (Al) -?
1) 6HCl + 2Al => 2AlCl3 + 3H2 ↑;
2) m (HCl) = ω (HCl) * m solution (HCl) / 100% = 25% * 2.7 / 100% = 0.675 g;
3) n (HCl) = m (HCl) / M (HCl) = 0.675 / 36.5 = 0.018 mol;
4) n (H2) = n (HCl) * 3/6 = 0.018 * 3/6 = 0.009 mol;
5) V (H2) = n (H2) * Vm = 0.009 * 22.4 = 0.2 l;
6) n (Al) = n (HCl) * 2/6 = 0.018 * 2/6 = 0.006 mol.
Answer: The volume of H2 is 0.2 l; the amount of Al substance is 0.006 mol.
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