Answer is: 6.66 L of NH3 are needed to react completely.
Balanced chemical equation: 4NH₃ + 6NO → 5N₂ + 6H₂O.
V(NO) = 10.0 L; volume of nitrogen(II) oxide.
V(NH₃) = ?
Vm = 22.4 L/mol; molar volume at STP.
n(NO) = V(NO) ÷ Vm.
n(NO) = 10 L ÷ 22.4 L/mol.
n(NO) = 0.446 mol; amount of nitrogen(II) oxide.
From balanced chemical equation: n(NO) : n(NH₃) = 6 : 4
0.446 mol : n(NH₃) = 6 : 4.
n(NH₃) = 0.3 mol; amount of ammonia.
V(NH₃) = 0.3 mol · 22.4 L/mol.
V(NH₃) = 6.66 L.
