Potassium trioxonitrate(V) decomposes upon heating according to the balanced chemical equa… — JAMB UTME

Potassium trioxonitrate(V) decomposes upon heating according to the balanced chemical equation:

2KNO_3(s) \rightarrow 2KNO_2(s) + O_2(g)

What volume of oxygen gas measured at STP is produced by the complete thermal decomposition of

50.5\text{ g}

KNO_3

[K = 39, N = 14, O = 16\text{; Molar gas volume at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]

The correct volume of oxygen gas produced at STP is

5.6\text{ dm}^3

The complete decomposition of

50.5\text{ g}

(

0.5\text{ mol}

) of

KNO_3

yields

0.25\text{ mol}

O_2

gas according to the

2:1

stoichiometric mole ratio. Multiplying

0.25\text{ mol}

by the standard molar gas volume at STP (

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

) gives

5.6\text{ dm}^3

Calculate the molar mass of

KNO_3

Molar mass of

KNO_3 = 39 + 14 + (3 \times 16) = 101\text{ g mol}^{-1}

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

Calculate the number of moles of

KNO_3

reacted

\text{Moles of } KNO_3 = \frac{50.5\text{ g}}{101\text{ g mol}^{-1}} = 0.5\text{ mol}

Determines the exact mole amount of reactant supplied.

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

O_2

produced

\text{Moles of } O_2 = \frac{1}{2} \times 0.5\text{ mol} = 0.25\text{ mol}

The equation shows that

2\text{ moles}

KNO_3

yield

1\text{ mole}

O_2

Convert moles of

O_2

to volume at STP

\text{Volume of } O_2 = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.6\text{ dm}^3

Molar gas volume at standard temperature and pressure (STP) is

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

Mass-Volume Stoichiometric Calculations at STP

Estimated Time:1m 30s

Question