Question

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

Answer 1

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 $O_{2}$