How many grams of #CO_2# are produced by the combustion of 484g of a mixture that is 35.1% #CH_4# and 64.9% #C_3H_8# by mass?
This reaction will produce 1410 g of carbon dioxide.
The trick here is to realize that you've got two combustion reactions taking place at the same time.
More specifically, you're going to have to write two balanced chemical equations, one for the combustion of methane and one for the combustion of propane.
The first thing you need to do is figure out exactly how many grams of each hydrocarbon you have. To do that, use the known percent composition of methane and propane in the mixture.
Check to see if the values add up to give 484 g.
Now focus on the two combustion reactions. Start with the first one
Notice that you have a
Use methane's molar mass to determine how many moles react
This means that this reaction will produce 10.59 moles of
Now for the second reaction
This time, the mole ratio that exists between propane and carbon dioxide is actually equal to
This means that you have
The total number of moles of carbon dioxide will be
Now simply use carbon dioxide's molar mass to determine how many grams would contain this many moles
Rounded to three sig figs, the answer will be