Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
