7) Answer is: c. 4.24 L.
Balanced chemical reaction: 4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g).
m(O₂) = 10.6 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 10.6 g ÷ 32 g/mol.
n(O₂) = 0.33 mol; amount of oxygen.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
V(O₂) = n(O₂) · Vm.
V(O₂) = 0.33 mol · 22.4 L/mol.
V(O₂) = 7.42 L; volume of oxygen.
From balanced chemical reaction: n(O₂) : n(NH₃) = 7 : 4.
n(NH₃) = 4 · 0.33 mol ÷ 7.
n(NH₃) = 0.188 mol; amount of ammonia.
V(NH₃) = 0.188 mol · 22.4 L/mol.
V(NH₃) = 4.24 L.
8) Answer is: a. 3.7 g.
Balanced chemical reaction: P₄(g) + 6H₂(g) → 4PH₃(g).
m(P₄) = 3.4 g; mass of phosphorous.
n(P₄) = m(P₄) ÷ M(P₄).
n(P₄) = 3.4 g ÷ 123.9 g/mol.
n(P₄) = 0.0274 mol; limiting reactant.
n(H₂) = 4 g ÷ 2 g/mol.
n(H₂) = 2 mol; amount of hydrogen.
From balanced chemical reaction: n(P₄) : n(PH₃) = 1 : 4.
n(PH₃) = 4 · 0.0274 mol.
n(PH₃) = 0.1096 mol.
m(PH₃) = 0.1096 mol · 34 g/mol.
m(PH₃) = 3.726 g.
9) Answer is: d. Percent yield is the ratio of theoretical yield to actual yield expressed as a percent.
Percent yield = actual yield / theoretical yield.
Theoretical yield is the maximum amount of product that can be produced from limiting reactant and actual yield is a product that is obtained by experimentation.
For example:
the percent yield = 250 g ÷ 294.24 g · 100%.
the percent yield = 84.5 %.
