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Question

What volume of hydrogen gas will be produced by reacting 15.5g of potassium in water at standard conditions of temperature and pressure? 2K(aq)+2H2O(aq)⟶2KOH(aq)+H2(g)

Answer 1

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Potassium + water ⟶ Potassium hydroxide + hydrogen

2K(aq) + 2H₂O(aq) ⟶ 2KOH(aq) + H₂(g)

K is 39.1 g/mol

15.5 g / 39.1 g/mol = 0.396 mol

mole ratio of H₂ to K is 1/2

therefore 0.396 mol will produce 0.396/2 = 0.198 mol of H₂

One mole of any ideal gas at STP has a volume of 22.41L

  (old def of STP, 1 atm)

using old def of STP, 22.41 L/mol x 0.198 mol = 4.44 L

Answer 2

(15.5 g K) / (39.0983 g K/mol) x (1 mol H2 / 2 mol K) x (22.4 L/mol) = 4.44 L H2