Propane burns completely in oxygen gas according to the following balanced chemical equati… — JAMB UTME

Propane burns completely in oxygen gas according to the following balanced chemical equation:

C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

What volume of oxygen gas, measured at STP, is required for the complete combustion of

4.4\text{ g}

of propane?

[\text{H} = 1.0, \text{C} = 12.0; \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1}]

The volume of oxygen gas required at STP is

11.2\text{ dm}^3

The option stating

11.2\text{ dm}^3

is correct.

4.4\text{ g}

of propane corresponds to

0.10\text{ mol}

. According to the balanced equation,

1\text{ mol}

C_3H_8

reacts with

5\text{ mol}

O_2

, so

0.10\text{ mol}

of propane requires

0.50\text{ mol}

O_2

. At STP,

0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

Calculate the molar mass of propane (

C_3H_8

Molar mass of

C_3H_8 = (3 \times 12.0) + (8 \times 1.0) = 36.0 + 8.0 = 44.0\text{ g mol}^{-1}

Molar mass is needed to convert the given mass of reactant into moles.

Determine the number of moles of propane in

4.4\text{ g}

Moles of

C_3H_8 = \frac{4.4\text{ g}}{44.0\text{ g mol}^{-1}} = 0.10\text{ mol}

Stoichiometric calculations rely on mole ratios from the balanced chemical equation.

Use the mole ratio from the balanced equation to find the required moles of

O_2

Mole ratio

C_3H_8 : O_2 = 1 : 5

. Moles of

O_2 = 0.10\text{ mol} \times 5 = 0.50\text{ mol}

Every

1\text{ mole}

of propane requires

5\text{ moles}

of oxygen gas for complete combustion.

Calculate the volume of

O_2

gas at STP.

Volume of

O_2 = 0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

Multiply the calculated number of moles of gas by the molar volume at STP (

22.4\text{ dm}^3\text{ mol}^{-1}

Mass-Volume Stoichiometric Calculations at STP

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