Answer to Question #123210 in General Chemistry for Nini

Question

C3H8 + 5 O2 - - - - - - - - > 3 CO2 + 4 H2O
Calculate the mass of CO2 produced when 6.5 g of propane is reacted with 14.2 g of O2.

Answer 1

First step is to determine the limiting reactant:

n (C₃H₈) = 6.5g / 44.1 g/mol = 0.147 mol

n (O₂) = 14.2g / 32 g/mol = 0.444 mol

According to the equation, the C₃H₈ : O₂ mole ratio is 1 : 5.

The actual ratio is 0.147 : 0.444 = 1 : 3 => O₂ is the limiting reactant.

Now moles and mass of CO₂ can be calculated.

n (CO₂) = 3/5 x n(O₂) = 0.267 mol

m (CO₂) = 0.267 mol x 44 g/mol = 11.7 g

Answer: 11.7 g