Question

During the combustion of methane (CH4), 35.2 g of methane reacts with 45.3 g of oxygen. What is the limiting reactant and theoretical yield for the formation of CO2?

Answer 1

Balanced chemical equation is

CH4 + 2 O2 ---------------> CO2 + 2 H2O

no. of moles of CH4 = 35.2/16 = 2.2 moles

no. of moles of O2 = 45.3 / 32 = 1.415 moles

from the equation, each mole of CH4 will react with 2 moles of O2

To react with 1.415 moles of O2, CH4 needed = (1/2) x 1.415 = 0.7078 moles

Here, CH4 present in excess

So, limiting reagent is O2

2 moles of O2 will give 1 mole of CO2

1.415 moles of O2 will give CO2 = (1/2) x 1.415 = 0.7078 moles of CO2

Weight of CO2 formed = 0.7078 x 44 = 31.14 grams