To determine the limiting reactant, we need to calculate how much CO2 and H2O each reactant can produce based on their given masses and stoichiometry.
For C2H6:
Molar mass of C2H6 = 2(12.01) + 6(1.01) = 30.07 g/mol
Number of moles of C2H6 = 21 g / 30.07 g/mol = 0.70 mol
From the balanced equation, 2 moles of C2H6 can produce 4 moles of CO2
Moles of CO2 that can be produced = 2/2 x 0.70 mol = 1.40 mol
From the balanced equation, 2 moles of C2H6 can produce 6 moles of H2O
Moles of H2O that can be produced = 2/2 x 0.70 mol = 1.40 mol
For O2:
Molar mass of O2 = 2(16.00) = 32.00 g/mol
Number of moles of O2 = 32 g / 32.00 g/mol = 1.00 mol
From the balanced equation, 1 mole of O2 can produce 2 moles of CO2
Moles of CO2 that can be produced = 2/7 x 1.00 mol = 0.29 mol
From the balanced equation, 1 mole of O2 can produce 3 moles of H2O
Moles of H2O that can be produced = 3/7 x 1.00 mol = 0.43 mol
Since O2 can only produce 0.29 mol of CO2 while C2H6 can produce 1.40 mol of CO2, O2 is the limiting reactant. Therefore, it will limit the amount of CO2 and H2O that can be produced in the reaction.
In the balanced equation
2C₂H6+702--> 4CO2+6H₂O
if 21 g of C₂H6 react with 32 g O2, what is the limiting reactant?
02
C₂H6
CO₂
H₂O
1 answer