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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