How much AL(OH3) is needed to make 1000g aluminium sulphate?

Question

the equation balanced is (sorry about big numbers, they dont do small ones)

2AL(OH)3 + 3H2SO4 = AL2(SO4)3 + 6H2O

This is for As level chemistry. I had a go but the number appeared to be too big

Answer 1

The balanced equation tells you that 2 moles of Al(OH)3 are required to produce 1 mole of Al2(SO4)3

Molar mass of Al2(SO4)3 is 342.16 g
1000 g = 2.9226 moles

You will need 5.845 moles of Al(OH)3 (2x2.9226)
Molar mass of Al(OH)3 = 78.00
78.00 g/mole x 5.845 moles = 455.9 g Al(OH)3

Answer 2

Remember to convert every thing to moles, and finally convert back to grams:
The molar mass of Al2(SO4)3 is 342.15g
The molar mass of Al(OH)3 is 78.00344g
That means 1000g aluminum sulphate is 1000/342.15 = 2.9227 Mol. As indicated by the reaction, 2*2.9227 Mol = 5.8454 Mol of Al(OH)3 is needed to make 1000g or 2.9227 Mol aluminum sulphate. That is 5.8454*78.00344 = 455.96g Al(OH)3