When 0.400 mole of potassium reacts with excess water at standard temperature and pressure as shown in the equation above, the volume of hydrogen gas produced is: 2 k (s) 2 h2o (l) → 2 k (aq) 2 oh - (aq) h2 (g) ?

The balanced ionic equation for the above reaction is as follows;

2K (s) + 2H₂O (l) - - - > 2K⁺ (aq) + 2OH⁻ (aq) + H₂ (g)

stoichiometry of K to H₂ is 2:1

K is the limiting reactant as H₂O is in excess.

number of moles of k reacted - 0.400 mol

according to molar ratio,

number of H₂ moles formed - 0.400 / 2 = 0.200 mol

at Standard temperature and pressure conditions,

molar volume is where 1 mol of any gas occupies a volume of 22.4 L

1 mol of H₂ occupies volume of 22.4 L

therefore 0.200 mol of H₂ occupies a volume of - 22.4 L/mol x 0.200 mol

volume of H₂ is = 4.48 L