The limiting reactant is H₂. C₂H₂ is in 43 % molar excess.
Step 1. Write the balanced chemical equation
C₂H₂ +2H₂ ⟶ C₂H₆
n: 1 1.4
Step 2. Identify the limiting reactant
Calculate the moles of C₂H₆ we can obtain from each reactant.
From C₂H₂: Moles of C₂H₆ = 1 mol C₂H₂ × (1 mol C₂H₆/1 mol C₂H₂) = 1 mol C₂H₆
From H₂: Moles of S = 1.4 mol H₂ × (1 mol C₂H₆/2 mol H₂) = 0.7 mol C₂H₆
H₂ is the limiting reactant because it gives the smaller amount of C₂H₆.
Step 4. Calculate the molar excess of C₂H₂.
Theoretical moles of C₂H₂ reacted = 1.4 mol H₂ × (1 mol C₂H₂/2 mol H₂)
= 0.7 mol C₂H₂
Excess moles of C₂H₂ = (1 – 0.7) mol = 0.3 mol
% Molar excess = excess moles/theoretical moles × 100 % = 0.3/0.7 × 100 %
= 43 %
