Question

Consider the balanced equation for the combustion of propane, C3H8 C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) If propane reacts with oxygen as above a. what is the limiting reagent in a mixture containing 5.00 g of C3H8 and 10.0 g of O2? b. what mass of the excess reagent remains after the reaction ? c. what mass of CO2 is formed when 1.00 g of C3H8 reacts completely?

Answer 1

Ans :

a) 5.00 g of propane = 5.00 / 44.1 = 0.11 moles

10.0 g of O2 = 10.0 / 32 = 0.3125 moles

1 mole of propane requires 5 moles of O2

So 0.11 moles of O2 will require 0.11 x 5 = 0.55 moles of O2

SO clearly O2 is the limiting reagent.

b) 0.3125 moles of O2 will utilise 0.3125 / 5 = 0.0625 moles of propane

Mass of propane utilised = 0.0625 x 44.01 = 2.75 grams

So the amount of propane in excess = 5.00 - 2.75

= 2.25 g

c) 1.00 g of C3H8 = 1.00 / 44.01 = 0.023 moles

1 mole of propane makes 3 moles of CO2

So 0.023 moles of propane will make 0.023 x 3 = 0.069 moles of CO2

Amount of CO2 formed = 0.069 x 44.1 = 3.0429 g