Propane burns in excess oxygen according to the following reaction.

C3H8 + 5O2 → 3CO2 + 4H2O

a. H ow many moles each of CO2 and H2O are formed from 3.85 mol of propane?

ans: 11.6 mol CO2, 15.4 mol H2O

b. If 0.647 mol of oxygen is used in the burning of propane, how many moles each of CO2 and H2O are produced? How many moles of C3H8 are consumed?

ans: 0.388 mol CO2, 0.518 mol H2O, 0.129 mol C3H8

Explanation:

A. Solve the answer using stoichiometry.  This means that you have to look at the coefficients of the equation to solve this.

The problem gave the moles of propane used to form product.  Your job is to find the moles of product formed.  Use the equation to solve this.  The coefficients represent moles.

Propane has no coefficient implying that the number in front of propane would be 1.  CO₂ has a coefficient of 3, and H₂O has a coefficient of 4.  This means that for every mole of propane, 3 moles of CO₂ and 4 moles of H₂O are formed.

Multiply the moles of propane by the coefficients of the products.

3.85 × 3 = 11.55 mol CO₂

3.85 × 4 = 15.4 mol H₂O

You have solved for moles of each product.

B. This problem has about the same concept as the one before.  Look at the equation.  For every 5 moles of O₂, 3 moles of CO₂ and 4 moles of H₂O are formed.  For every 5 moles of O₂, 1 mole of propane is needed.

This time, however, you cannot simply multiply the amount of moles by the coefficient.  This is because you need 5 moles, not 1, to get that many moles produced.

This time, multiply the moles by the coefficients over the original coefficient.

0.647 × 3/5 = 0.388 mol CO₂

0.647 × 4/5 = 0.518 mol H₂O

0.647 × 1/5 = 0.129 mol C₃H₈

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