Question

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)?

Answer 1

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