How many grams of C3H8 are needed to react with excess oxygen to produce 156 grams of water? C3H8+5O2=3CO2+4H2O

1 Answer
SCooke
Apr 25, 2018

95.3g C_3H_8

Explanation:

First, check to ensure that the equation is balanced (it is). Then calculate the required moles of product water for a mass of 156 grams. Then use the ratio of moles of product H_2O to reactant C_3H_8 to find the moles of propane required. Finally, use that value with the molecular weight of propane (44) to find the mass of propane required.

156/18 = 8.67 moles H_2O

(1 "mol" C_3H_8)/(4"mol" H_2O) xx 8.67 = 2.17 moles Propane

2.17 xx 44 = 95.3g

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