Question

Calculate the amount of CO2 produced when 1.00 mole of propane (C3H8) is burned in the presence of 1.00 mole of oxygen.

Answer 1

Answer:

The balanced Chemical equation is

$1C_3 H_8  +5O_2  >3CO_2+4H_2 O$

Since quantity of two reactants are given Hence we find the moles of product using both.

The quantity of product produced is the one we consider from the limiting reactant.

Limiting reactant is the reactant which runs out first.

So, least product is produced from the limiting reactant.

Mole ratio of $C_3 H_8:CO_2$ is 1: 3

$1.00 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{\mathrm{8}} \times \frac{3 \mathrm{mol} \mathrm{CO}_{2}}{1 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8}}=3 \mathrm{molCO}_{2}$

Mole ratio of $O_2:CO_2$ is 5: 3

$1.00 \mathrm{mol} \mathrm{O}_{2} \times \frac{3 \mathrm{mol} \mathrm{co}_{2}}{5 \mathrm{mol} \mathrm{O}_{2}}=0.6 \mathrm{mol} \mathrm{CO}_{2}$

(Least produced)

So, the limiting reactant is $O_2$

The amount of $CO_2$ formed is 0.600mol (Answer)