Question

If 10.0 g of sodium metal is dropped into water, how much hydrogen gas will be produced?

Answer 1

Sodium reacts with water according to following reaction
2Na(s) + 2H2O → 2NaOH(aq) + H2(g)
Moles of sodium metal = 10.0 g/ 23 g/mole = 0.43 moles Na
Correspondingly moles of H2(g) produced = 0.43/2 = 0.22 moles H2(g)
Mass of hydrogen produced = 0.22 moles x 2 g/mole = 0.44 g H2(g)