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For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

A2.72 g KClO3
B0.361 g KClO3
C83.6 kg KClO3
D22.5 mg KClO3

Verified Answer :

Explanation:

Molar mass of \(KClO_{3}\) = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of \(O_{2}\) = 32.0 g

According to the equation, 2 moles of \(KClO_{3}\) reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of \(KClO_{3}\) will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of \(KClO_{3}\) gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of \(KClO_{3}\) as follows.

       \(\frac {96g}{245.2g} \times 2.72 g = 1.06 g\) of \(O_{2}\)


(b) Calculate the amount of oxygen given by 0.361 g of \(KClO_{3}\) as follows.

    \(\frac {96g}{245.2g} \times 0.361 g = 0.141 g\) of \(O_{2}\)


c) Calculate the amount of oxygen given by 83.6 kg \(KClO_{3}\) as follows.  

    \(\frac {96g}{245.2g} \times 83.6 g = 32.731 kg\) of \(O_{2}\)

Convert kg into grams as follows.

    \(\frac{32.731 kg}{1 kg} \times 1000 g\) = 32731 g of \(O_{2}\)


(d) Calculate the amount of oxygen given by 22.5 mg of \(KClO_{3}\) as follows.  

    \(\frac {96g}{245.2g} \times 22.5 mg = 8.79 mg\)

Convert mg into grams as follows.

  \(\frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g\) of \(O_{2}\)

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