Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s).

The equation for the reaction is

2KClO3⟶2KCl+3O2

Calculate how many grams of O2(g) can be produced from heating 77.7 g KClO3(s).

Answer 1

30.4 g of O2

Step-by-step explanation:

To find the number of moles of KClO3, use the formula: moles = mass / molar mass.

Molar mass of KClO3 is = (39.1+35.5+3*16)=122.5g/mol

moles of KClO3 = 77.7g / 122.5g/mol = 0.634 mol

Therefore, the number of moles of O2 that can be produced is (3 moles O2/ 2 moles KClO3) x 0.634 moles KClO3 = 0.951 moles O2.

To find the mass of O2 produced, we use the formula: mass = moles x molar mass

molar mass of O2 = 2*16 =32 g/mol

mass of O2 = 0.951 moles * 32g/mol = 30.4 g of O2

So, from heating 77.7 g KClO3(s) 30.4 g of O2 can be produced.