How many grams of #O_2(g)# are needed to completely burn #21.4# #g# of #C_3H_8(g)# in the reaction #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#?

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How many grams of #O_2(g)# are needed to completely burn #21.4# #g# of #C_3H_8(g)# in the reaction #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#?

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Answer 1 · 2016-03-22T23:51:56

Find the number of moles of #C_3H_8#: #n=m/M=21.4/44=0.49# #mol#. Each mole of #C_3H_8# requires #5# #mol# of #O_2#, so we need #5xx0.49=2.45# #mol#. The mass will be #m=nM=2.45xx32=78.4# #g#.

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How many grams of #O_2(g)# are needed to completely burn #21.4# #g# of #C_3H_8(g)# in the reaction #C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)#?

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