Consider the balanced chemical equation for the complete combustion of ethane gas: 2C_2H_6… — JAMB UTME

Consider the balanced chemical equation for the complete combustion of ethane gas:

2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)

What volume of oxygen gas, measured at STP, is required to react completely with

0.5\text{ mol}

of ethane? Complete the statement below with the calculated numerical value.

Answer:The volume of oxygen gas required at STP is 【39.2】

\text{dm}^3

. (Molar gas volume at STP =

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

)

39.2 dm³ (or 39.2)

According to the balanced equation

2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)

2\text{ moles}

of ethane require

7\text{ moles}

of oxygen gas for complete combustion. Therefore,

0.5\text{ mol}

of ethane requires

\frac{0.5 \times 7}{2} = 1.75\text{ mol}

of oxygen gas. At standard temperature and pressure (STP),

1\text{ mole}

of gas occupies

22.4\text{ dm}^3

. Thus, the volume of oxygen gas required is

1.75\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 39.2\text{ dm}^3

Identify the stoichiometric mole ratio between ethane and oxygen from the balanced equation

From

2C_2H_6 + 7O_2

, the mole ratio of

C_2H_6

O_2

2 : 7

Stoichiometric coefficients represent the relative mole proportions of reactants.

Calculate the moles of oxygen gas needed for 0.5 mol of ethane

Moles of

O_2 = 0.5\text{ mol} \times \frac{7}{2} = 1.75\text{ mol}

Multiplying the given amount of ethane by the stoichiometric factor gives the required moles of oxygen.

Convert the calculated moles of oxygen gas to volume at STP

Volume of

O_2 = 1.75\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 39.2\text{ dm}^3

One mole of any ideal gas occupies

22.4\text{ dm}^3

at STP.

Mole-Volume Stoichiometric Calculations at STP

Question