Question

How many moles are there in
15 grams of:
CaCO3
204 grams of:
Al203

Answer 1

1) moles = mass/mR

CaCO3 Mr = 40 + 12 + (16×3)

= 52 + 48

= 100

mass = 15

so the moles would be 15 ÷ 100

which is 0.15 moles of CaCO3

2) moles = mass ÷ Mr

Mr of Al2O3 = 27 + (16×3)

= 27 + 48

= 75

mass = 204

so the moles would be 204/75 which is 2.72 moles of Al2O3