A 3.50\text{ g} sample of impure potassium trioxochlorate(V), \text{KClO}_3, was completel… — JAMB UTME

3.50\text{ g}

sample of impure potassium trioxochlorate(V),

\text{KClO}_3

, was completely decomposed by heating in the presence of a manganese(IV) oxide catalyst according to the equation:

2\text{KClO}_3(\text{s}) \rightarrow 2\text{KCl}(\text{s}) + 3\text{O}_2(\text{g})

0.672\text{ dm}^3

of oxygen gas was collected at STP, what is the percentage purity of the

\text{KClO}_3

sample?

(\text{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})

The percentage purity of the potassium trioxochlorate(V) sample is 70.0%.

The option stating 70.0% is correct. Converting

0.672\text{ dm}^3

\text{O}_2

gas at STP yields

0.030\text{ mol}

\text{O}_2

. Using the balanced stoichiometric mole ratio (

2\text{ KClO}_3 : 3\text{ O}_2

), this corresponds to

0.020\text{ mol}

of pure

\text{KClO}_3

. Multiplying by its molar mass (

122.5\text{ g/mol}

) gives

2.45\text{ g}

of pure

\text{KClO}_3

. Dividing

2.45\text{ g}

by the total sample mass of

3.50\text{ g}

and multiplying by

100\%

gives

70.0\%

Calculate the moles of oxygen gas collected at STP

\text{Moles of O}_2 = \frac{0.672\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.030\text{ mol}

At STP, one mole of any ideal gas occupies 22.4 dm³.

Determine the moles of pure potassium trioxochlorate(V) reacted using stoichiometric coefficients

\text{Moles of KClO}_3 = 0.030\text{ mol O}_2 \times \frac{2\text{ mol KClO}_3}{3\text{ mol O}_2} = 0.020\text{ mol}

From the balanced chemical equation, 2 moles of KClO3 decompose to yield 3 moles of O2.

Calculate the mass of pure potassium trioxochlorate(V)

\text{Molar mass of KClO}_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g/mol}

\text{Mass of pure KClO}_3 = 0.020\text{ mol} \times 122.5\text{ g/mol} = 2.45\text{ g}

Mass is obtained by multiplying moles by molar mass.

Calculate the percentage purity of the original sample

\text{Percentage purity} = \left(\frac{2.45\text{ g}}{3.50\text{ g}}\right) \times 100\% = 70.0\%

Percentage purity is the ratio of pure substance mass to total sample mass expressed as a percentage.

Percentage purity calculation via gas stoichiometry at STP

Question