Question

How many grams of CO2 are produced by the combustion of 289 g of a mixture that is 28.6% CH4 and 71.4% C3H8 by mass?

Answer 1

845 grams of CO2 are produced by the combustion of 289 g of a mixture that is 28.6% CH4 and 71.4% C3H8 by mass.

Explanation:

The mixture of methane and propane is butane with molecular formula

C4H12

on combustion the balanced reaction is:

C4H12 + 7 O2 ⇒   4CO2  + 6  H20

Molar mass of butane = 60.13 grams/mole, mass of C4H12 = 289 grams

number of moles of C2H12 is calculated by using the formula:

n = $\frac{mass}{molar mass of 1 mole}$

putting the values to know the moles of C4H12:

n = $\frac{289}{60.13}$

n = 4.8 moles of C4H12

from the balanced equation:

1 mole of C4H12 undergoes combustion to form 4 moles of CO2

thus, 4.8 moles when undergoes combustion will form x moles of CO2

$\frac{4}{1}$= $\frac{x}{4.8}$

X = 4 X 4.8

x = 19.2 moles of CO2 is formed.

to convert mole into mass, formula used:

mass = number of moles x atomic mass

         = 19.2 x 44.01

          = 844.99 grams of CO2 is formed.