Answer:
1) Ethane is the limiting reactant
2) mass of CO₂ produced is 733.33 g
Explanation:
C₂H₆ + 2O₂ → 2CO₂ + 3H₂ ------------------------(1)
molar ratio for equation (1) can be shown as;
1 : 2 → 2 : 3
molecular weight of moles of C₂H₆ = 30 g/mol
molecular weight of moles of O₂ = 32 g/mol
molecular weight of moles of CO₂ = 44 g/mol
mass of O₂ = 220 g
mass of C₂H₆ = 250 g
number of moles of O₂ = mass ÷ molecular weight = 220 g ÷ 32 g/mol * 2 = 13.75 moles
number of moles of C₂H₆ = mass ÷ molecular weight = 250 g ÷ 30 g/mol = 8.333 moles
1) Ethane is the limiting reactant as the 8.333 moles will finish before the 13.75 moles of the oxygen is totally consumed.
2) 1 mole of C₂H₆ will form 2 moles of CO₂
Therefore, 8.333 moles will form 16.667 moles of CO₂
mass of CO₂ formed = number of moles * molecular weight = 16.667 * 44 = 733.33 g
