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 = [tex]\frac{mass}{molar mass of 1 mole}[/tex]
putting the values to know the moles of C4H12:
n = [tex]\frac{289}{60.13}[/tex]
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
[tex]\frac{4}{1}[/tex]= [tex]\frac{x}{4.8}[/tex]
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.
