Question

A sample of propane, C3H8, with a mass of 9.61 g is completely combusted in an excess of oxygen under room conditions.

Which volume of carbon dioxide gas is produced?

A 4.89 dm3 B 5.24 dm3 C 14.7 dm3 D 15.7 dm3

Answer 1

Answer:

A sample of propane, C3H8, with a mass of 9.61 g is completely combusted in an excess of oxygen under room conditions.

Which volume of carbon dioxide gas is produced?

A 4.89 dm3 B 5.24 dm3 C 14.7 dm3 D 15.7 dm3

Explanation:

A sample of propane, C3H8, with a mass of 9.61 g is completely combusted in an excess of oxygen under room conditions.

Which volume of carbon dioxide gas is produced?

A 4.89 dm3 B 5.24 dm3 C 14.7 dm3 D 15.7 dm3

Answer 2

Given: Mass of given sample of propane C₃H₈ = 9.61 g

           This sample is completely combusted in excess of oxygen

To Find : Volume of CO₂ evolved

Solution:

• Let us write balanced equation for the process of combustion:

                 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

From this equation it is clear that 1 mole of propane reacts with 5 moles of oxygen to evolve 3 moles of carbon dioxide and 4 moles of water.

• Now we will calculate the number of moles of propane:

Molecular mass of C₃H₈ = 3(12) + 8(1) = 44 g

   If 44 g of propane = 1 mole

∴ 9.61 g of propane = $\frac{1}{44}$ × 9.61 = 0.218 moles

• Now we will calculate number of moles of carbon dioxide evolved from 0.218 moles of propane:

   1 mole of propane gives 3 moles of carbon dioxide

∴ 0.218 moles of propane will evolve 3 × 0.218 = 0.654 moles of CO₂

• Now we will calculate volume of CO₂ evolved.

        1 mole of CO₂ = 22.4 L

       ∴ 0.654 moles of CO₂ = 22.4 × 0.654 = 14.64 L ≈ 14.7 L

Hence on complete combustion of 9.61 g of propane 14.7 $dm^{3}$ carbon dioxide is evolved.

So option (C) 14.7 $dm^{3}$ is correct.