Question

How many grams of octane (C8H18) are needed to produce 100.0 g of carbon dioxide in a complete combustion reaction if the reaction occurs with a 60.0% yield?

Answer 1

for CO2

% yield = actual mass*100/theoretical mass

60.0= 100*100/theoretical mass

theoretical mass = 166.7 g

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = 166.7 g

mol of CO2 = (mass)/(molar mass)

= 166.7/44.01

= 3.7878 mol

we have the Balanced chemical equation as:

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

From balanced chemical reaction, we see that

when 16 mol of CO2 forms, 2 mol of C8H18 is reacting

mol of C8H18 reacting = (2/16)* moles of CO2

= (2/16)*3.7878

= 0.4735 mol

Molar mass of C8H18 = 8*MM(C) + 18*MM(H)

= 8*12.01 + 18*1.008

= 114.224 g/mol

mass of C8H18 = number of mol * molar mass

= 0.4735*114.224

= 54.1 g

Answer: 54.1 g