Oxygen is prepared by catalytic decompostion of potassium chlorate, KClO3. Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2). If 2.4 mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?

Question

Oxygen is prepared by catalytic decompostion of potassium chlorate, KClO3. Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2). If 2.4 mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed? (A) 196.0 g (B) 120 g (C) 56.0 g (D) 224 g

Tardigrade Admin · Accepted Answer

2KClO3(s) → 2KCl(s) + 3O2(g) ∴ 3 mol of oxygen requires 2 mol of KClO3. For 2.4 mol of oxygen, we need = 1.6 mol of KClO3 Therefore, mass of KClO3 required = 1.6 mol × 122.5 g mol-1 (M.W. of KClO3 = 122.5) = 196.0 g

Q. Oxygen is prepared by catalytic decompostion of potassium chlorate, $K Cl O_{3}$ . Decomposition of potassium chlorate gives potassium chloride $(K Cl)$ and oxygen $(O_{2})$ . If $2.4$ mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?

196.0 g

120 g

56.0 g

224 g

$2 K Cl O_{3 (s)} \to 2 K C l_{(s)} + 3 O_{2 (g)}$
$∴ 3$ mol of oxygen requires $2$ mol of $K Cl O_{3}$ .
For $2.4$ mol of oxygen, we need $= 1.6 m o l$ of $K Cl O_{3}$
Therefore, mass of $K Cl O_{3}$ required =
$1.6 m o l \times 122.5 g m o l^{- 1} (M . W .$ of $K Cl O_{3} = 122.5)$
$= 196.0 g$

Q. Oxygen is prepared by catalytic decompostion of potassium chlorate, KClO3​. Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2​). If 2.4 mole of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?