If 10.0 g of sodium metal is dropped into water, how much hydrogen gas will be produced?
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Expert's answer
2018-03-13T11:01:07-0400
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)
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