Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the reaction is 2 KClO 3 ⟶ 2 KCl + 3 O 2 Calculate how many grams of O 2 ( g ) can be produced from heating 82.4 g KClO 3 ( s ) .

Answer 1

Answer:

Mass of $O_2$ produced = 32 g

Explanation:

Calculation of the moles of $KClO_3$ as:-

Mass = 82.4 g

Molar mass of $KClO_3$ = 122.55 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{82.4\ g}{122.55\ g/mol}$

$Moles= 0.67237\ mol$

From the reaction shown below:-

$2KClO_3\rightarrow 2KCl+3O_2$

2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms $\frac{3}{2}\times 0.67237$ moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas  = 32 g/mol

Mass of $O_2$ produced = 32 g