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?

1 Answer
Jul 10, 2015

Answer:

This reaction will produce 1410 g of carbon dioxide.

Explanation:

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.

#484cancel("g mixture") * ("35.1 g"CH_4)/(100cancel("g mixture")) = "169.9 g"# #CH_4#

and

#484cancel("g mixture") * ("64.9 g" C_3H_8)/(100cancel("g mixture")) = "314.1 g"# #C_3H_8#

Check to see if the values add up to give 484 g.

#169.9 + 314.1 = "484 g"# #-># so far, so good.

Now focus on the two combustion reactions. Start with the first one

#CH_(4(g)) + 2O_(2(g)) -> CO_(2(g)) + 2H_2O_((l))#

Notice that you have a #1:1# mole ratio between methane and carbon dioxide. This means that, regardless of how many moles of methane eract, the reaction will produce the same exact number of moles of carbon dioxide.

Use methane's molar mass to determine how many moles react

#169.9cancel("g") * "1 mole"/(16.04cancel("g")) = "10.59 moles"# #CH_4#

This means that this reaction will produce 10.59 moles of #CO_2#.

Now for the second reaction

#C_3H_(8(g)) + 5O_(2(g)) -> color(red)(3)CO_(2(g)) + 4H_2O_((l))#

This time, the mole ratio that exists between propane and carbon dioxide is actually equal to #1:color(red)(3)#. So, for every mole of propane that reacts, the reaction produces 3 times more moles of #CO_2#.

#314.1cancel("g") * "1 mole"/(44.10cancel("g")) = "7.12 moles"# #C_3H_8#

This means that you have

#7.12cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole" C_3H_8)) = "21.36 moles"# #CO_2#

The total number of moles of carbon dioxide will be

#n_"total" = 10.59 + 21.36 = "31.95 moles"# #CO_2#

Now simply use carbon dioxide's molar mass to determine how many grams would contain this many moles

#31.95cancel("moles") * "44.01 g"/(1cancel("mole")) = "1406.1 g"# #CO_2#

Rounded to three sig figs, the answer will be

#m_(CO_2) = color(green)("1410 g "CO_2)#