Question

Calculate the mass of potassium chlorate required to liberate 6.72 ${\text{d}}{{\text{m}}^{\text{3}}}$ of oxygen at STP. Molar mass of potassium chlorate is 122.5 g/mol.

Answer 1

You can solve this problem using Avogadro’s law. Avogadro’s law says that at the same temperature and pressure, equal volumes of gases contain equal numbers of molecules.

Complete step by step answer:
We can solve this problem using Avogadro’s law. According to Avogadro’s law, one mole of every substance occupy 22.4 L at STP and contains Avogadro number of particles, that is {\text{6}}{\text{.022 \times 1}}{{\text{0}}^{{\text{23}}}} particles. The reaction is as follows:
$2KCl{O_3}(s) \to 2KCl(s) + 3{O_2}(g)$
$2KCl{O_3}(s) \to 2KCl(s) + 3{O_2}(g)$
According to stoichiometry:
2 moles of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ produce 3 moles of ${{\text{O}}_{\text{2}}}$.
Thus $3 \times 22.4L = 67.2L$ of ${{\text{O}}_{\text{2}}}$ is produced from $2 \times 122.5 = 245g$ of ${\text{KCl}}{{\text{O}}_{\text{3}}}$
6.72 L of ${{\text{O}}_{\text{2}}}$ is produced from = $\dfrac{{245}}{{10}}$ = 24.5 of ${\text{KCl}}{{\text{O}}_{\text{3}}}$
Thus mass of potassium chlorate required to liberate 6.72 ${\text{d}}{{\text{m}}^{\text{3}}}$of oxygen at STP will be 24.5 grams.

Additional Information: Stoichiometry is the branch of study in chemistry which involves using relationships between reactants and products to determine the desired quantitative data from a chemical reaction. The word ‘stoichiometry’ is formed from two Greek words ‘stoikhein’ and ‘metron’ having the meaning of ‘element’ and ‘measure’ respectively.

Note: When you are doing this kind of problem, in the first step you can write how much product is formed from a given amount of reactants. Then, find out the amount of reactants required for one gram of products. Then multiply this with the required amount of products.