What volume of hydrogen will be released when sodium weighing 2.3 g interacts with hydrochloric acid?

Reaction: 2Na + 2HCl = 2NaCl + H2
The amount of sodium substance n = m / M = 2.3 g / 23 g / mol = 0.1 mol. (M is the molar mass determined by the periodic table). By reaction, from 2 mol of sodium, 1 mol of hydrogen is obtained. 0.1 / 2 = n (H2) / 1. n (H2) = 0.1 / 2 = 0.05 mol. Find the volume of hydrogen V = n * Vm = 0.05 mol * 22.4 l / mol = 1.12 l. (Vm – molar volume, constant for all gases under normal conditions).
Answer: 1.12 liters of hydrogen will be released.




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