For the reaction 2kclo3 (s) →2kcl (s) + 3o2 (g) calculate how many grams of oxygen form when each quantity of reactant completely reacts.

First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl = 39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3:

when 245.2 KClO3 gives → 96 g O2

2.72 g KClO3 gives → X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

= 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

= 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

83.6 Kg KClO3 gives → X

∴X = 83.6 Kg * 96 g O2 / 245.2 g KClO3

= 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

= 8.8 mg of O2