Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.1 grams of ethane is mixed with 3.68 grams of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction.

= 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

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explanation: