What volume of gas will be released during the interaction of 2.3 g of Na with water

2Na + 2H2O = 2NaOH + H2

Let’s find the amount of sodium substance:

n (Na) = m (Na) / M (Na) = 2.3 / 23 = 0.1 mol;

According to the stoichiometry of the reaction:

n (H2) = 0.5n (Na) = 0.05 mol;

Let’s find the volume of hydrogen (at n.a.):

V (H2) = n (H2) * VM = 0.05 * 22.4 = 1.12 liters.

Answer: V (H2) = 1.12 l.

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