How many grams of CO2 are produced by the combustion of 309 g of a mixture...

Question

Question

How many grams of CO2 are produced by the combustion of 309 g of a mixture that is 43.6% CH4 and 56.4% C3H7 by mass? 5229 O 1

Answer 1

Answer #1

Answer

892g

Explanation

Mass of CH4 in the mixture = (43.6/100) × 309g = 134.724g

Mass of C3H8 in the mixture = ( 56.4/100) × 309g = 174.276g

number of moles of CH4 = 134.724g/16.05g/mol = 8.394mol

number of moles of C3H8 = 174.276g/44.11g/mol = 3.951mol

CH4(g) + 2O2(g) -----> CO2(g) + 2H2O(l)

C3H8(g) + 5O2(g) -------> 3CO2(g) + 4H2O(l)

number of moles of CO2 obtained by 8.394moles of CH4 = 8.394mol

number of moles of CO2 obtained by 3.951moles of C3H8 = 3× 3.951mol = 11.853mol

Total moles of CO2 = 8.394mol + 11.853mol = 20.247 mol

mass of CO2 produced = 20.247mol × 44.01g/mol = 891 g

Answer 2

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