Consider the thermal decomposition of potassium trioxochlorate(V) represented by the balan… — JAMB UTME

Consider the thermal decomposition of potassium trioxochlorate(V) represented by the balanced equation:

2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)

What mass of

KClO_3

is required to produce

6.72\text{ dm}^3

of oxygen gas measured at STP?

[K = 39.0,\text{ } Cl = 35.5,\text{ } O = 16.0;\text{ Molar volume of gas at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]

24.5\text{ g}

KClO_3

is required.

According to the balanced chemical equation,

2\text{ moles}

KClO_3

(

245.0\text{ g}

) produce

3\text{ moles}

O_2

(

67.2\text{ dm}^3

at STP). By direct proportion,

6.72\text{ dm}^3

O_2

requires

\frac{6.72}{67.2} \times 245.0 = 24.5\text{ g}

KClO_3

Calculate the molar mass of

KClO_3

and the total mass of 2 moles of

KClO_3

Molar mass of

KClO_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g mol}^{-1}

. Mass of

2\text{ moles} = 2 \times 122.5 = 245.0\text{ g}

The balanced chemical equation shows

2\text{ moles}

KClO_3

undergo decomposition.

Calculate the volume of 3 moles of

O_2

at STP.

Volume of

3\text{ moles of } O_2 = 3 \times 22.4\text{ dm}^3 = 67.2\text{ dm}^3

At STP,

1\text{ mole}

of any gas occupies

22.4\text{ dm}^3

Set up a proportion to find the mass of

KClO_3

needed to yield

6.72\text{ dm}^3

O_2

\text{Mass of } KClO_3 = \frac{6.72\text{ dm}^3}{67.2\text{ dm}^3} \times 245.0\text{ g} = 24.5\text{ g}.

Direct stoichiometric ratio relates mass of reactant to volume of gaseous product.

Mass-Volume Stoichiometry at STP

Question