Question

When propane (C3H8) is combusted with excess oxygen, carbon dioxide and water vapor are produced.

a. Determine how many grams of each reactant are required to produce 50.0 g of carbon dioxide.

b. If you combined the number of grams of both reactants determined in part a and produced 47.31 grams of carbon dioxide, what would the percent error for the experiment be?

Answer 1

a)

Balanced chemical equation is:

C3H8 + 5 O2 ---> 4 H2O + 3 CO2

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = 50 g

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

= 50/44.01

= 1.136 mol

According to balanced equation

mol of C3H8 required = (1/3)* moles of CO2

= (1/3)*1.136

= 0.3787 mol

Molar mass of C3H8,

MM = 3*MM(C) + 8*MM(H)

= 3*12.01 + 8*1.008

= 44.094 g/mol

mass of C3H8 = number of mol * molar mass

= 0.3787*44.09

= 16.7 g

According to balanced equation

mol of O2 required = (5/3)* moles of CO2

= (5/3)*1.136

= 1.894 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 1.894*32

= 60.6 g

mass of C3H8 = 16.7 g

mass of CO2 = 60.6 g

b)

error = 50.0 g = 47.31 g

= 2.69 g

% error = error * 100 / theoretical value

= 2.69*100/50.0

= 5.38 %

Answer: 5.38 %