Phosphorus reacts with oxygen gas to produce phosphorus(V) oxide according to the balanced… — JAMB UTME

Phosphorus reacts with oxygen gas to produce phosphorus(V) oxide according to the balanced chemical equation:

4\text{P}_{(s)} + 5\text{O}_{2(g)} \rightarrow \text{P}_4\text{O}_{10(s)}

If a mixture containing

12.4\text{ g}

of phosphorus and

20.0\text{ g}

of oxygen gas is allowed to react to completion, what is the mass of the excess reactant remaining unreacted in grams?

[\text{P} = 31.0, \text{O} = 16.0]

Answer: 4 g

Answer

The mass of unreacted excess oxygen gas remaining is

4.0\text{ g}

The correct calculation shows that

0.40\text{ mol}

of phosphorus requires

0.50\text{ mol}

of oxygen gas for complete reaction according to the

4:5

mole ratio in

4\text{P} + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}

. Subtracting the

0.50\text{ mol}

consumed from the initial

0.625\text{ mol}

leaves

0.125\text{ mol}

of unreacted oxygen gas, which equals

4.0\text{ g}

Step-by-Step Solution

Calculate the moles of phosphorus and oxygen gas present initially.

\text{Moles of P} = \frac{12.4\text{ g}}{31.0\text{ g/mol}} = 0.40\text{ mol}

;

\text{Moles of O}_2 = \frac{20.0\text{ g}}{32.0\text{ g/mol}} = 0.625\text{ mol}

Molar mass of

\text{P}

31.0\text{ g/mol}

and molar mass of

\text{O}_2

2 \times 16.0 = 32.0\text{ g/mol}

Determine the limiting reactant by comparing the required mole ratio to the available mole ratio.

\text{P}

is the limiting reactant, and

\text{O}_2

is the excess reactant.

From the balanced equation,

4\text{ moles of P}

require

5\text{ moles of O}_2

, so

1\text{ mole of P}

requires

1.25\text{ moles of O}_2

. Thus,

0.40\text{ mol}

\text{P}

requires

0.40 \times 1.25 = 0.50\text{ mol}

\text{O}_2

. Since

0.625\text{ mol}

\text{O}_2

is available,

\text{O}_2

is in excess.

Calculate the unreacted moles of oxygen gas remaining.

\text{Excess moles of O}_2 = 0.625\text{ mol} - 0.50\text{ mol} = 0.125\text{ mol}.

Subtracting the reacted moles from the initial moles gives the remaining amount.

Convert the unreacted moles of oxygen gas to mass in grams.

\text{Mass of remaining O}_2 = 0.125\text{ mol} \times 32.0\text{ g/mol} = 4.0\text{ g}.

Multiplying the excess moles by the molar mass of

\text{O}_2

(

32.0\text{ g/mol}

) gives the mass in grams.

Key Concept

Limiting and excess reactant stoichiometry

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