Question #ad087

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Answer 1 · 2016-12-15T04:31:19

13.6 L of #CO_2#

First, you need the balanced equation:

#C_3H_8# + 5 #O_2# ----> 3 #CO_2# + 4 #H_2O#

Convert 8.92 g into moles of propane (molar mass = 44 g)

8.92 / 44 = 0.2027 mol

According to the equation, 3 mol of #CO_2# is produced for each 1 mol of propane that reacts. So, in this problem, we obtain 0.608 mol of #CO_2#.

To find the volume at STP, multiply this by molar volume (22.4 L)

0.608 x 22.4 = 13.6 L