2C2H6 + 7O2 4CO2 + 6H2O If you are given 5.00 grams of C2H6 and an excess amount

Since we are told that O2 is the excess reactant, we need only to focus on C2H6, which will be our limiting reactant. Convert the mass of C2H6 to moles by dividing the mass by the molar mass of C2H6:

(5 g C2H6)(30.069 g/mol) = 0.1663 mol C2H6

Since C2H6 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and C2H6 per the balanced equation is 6:2. That is, for every 6 moles of H2O that is produced, 2 moles of C2H6 is used up (intuitively, then, the number of moles of H2O produced should be greater than the number of moles of C2H6 consumed, specifically 3 times greater).

So, the number of moles of H2O produced would be (0.1663 mol C2H6)(6 mol H2O/2 mol C2H6) = 0.499 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.499 mol H2O)(18.0153 g/mol) = 8.987 g H2O.

Note: If significant figures must be considered, then the answer would be 8.99 g H2O.

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