What is the mass of aluminum that will react with sodium hydroxide if 3.36 liters of gas are released?

Given:
V (gas) = 3.36 l

To find:
m (Al) -?

1) 2Al + 6NaOH => 2Na3AlO3 + 3H2 ↑;
2) n (H2) = V / Vm = 3.36 / 22.4 = 0.15 mol;
3) n (Al) = n (H2) * 2/3 = 0.15 * 2/3 = 0.1 mol;
4) m (Al) = n * M = 0.1 * 27 = 2.7 g.

Answer: The mass of Al is 2.7 g.



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