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?

Question

Naomi Aronson · Accepted Answer

This reaction will produce 1410 g of carbon dioxide.

Understanding that two combustion reactions are occurring simultaneously is the key in this situation. To be more precise, you will need to write two balanced chemical equations: one for propane combustion and one for methane combustion. The first step is to calculate the precise grams of each hydrocarbon, which can be done by using the mixture's known percent composition of methane and propane. #484cancel("g mixture") * ("35.1 g"CH_4)/(100cancel("g mixture")) = "169.9 g"# #CH_4# additionally #484cancel("g mixture") * ("64.9 g" C_3H_8)/(100cancel("g mixture")) = "314.1 g"# #C_3H_8# Verify if the total of the values equals 484 g. #169.9 + 314.1 = "484 g"# #-># so far, so good. The two combustion reactions are now your main focus; begin 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. Calculate how many moles of methane react by using its molar mass. #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#. Let's move on to the second response. #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 indicates that you've #7.12cancel("moles"C_3H_8) * (color(red)(3)" moles "CO_2)/(1cancel("mole" C_3H_8)) = "21.36 moles"# #CO_2# The total mass of carbon dioxide in moles will be #n_"total" = 10.59 + 21.36 = "31.95 moles"# #CO_2# Now just calculate how many grams would contain this many moles using the molar mass of carbon dioxide. #31.95cancel("moles") * "44.01 g"/(1cancel("mole")) = "1406.1 g"# #CO_2# The response, rounded to three sig figs, is #m_(CO_2) = color(green)("1410 g "CO_2)#

Natalie Anton · Answer

undefined The specified mixture when burned yields 1500 grams of CO2.