Question

A student performs a titration analysis for the following balanced reaction: 3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O If a 200.0g sample of Al(OH)3 is used to react with excess H2SO4, How many grams of aluminum sulfate are produced? a 157g b 630g c 1750g d 438g

Answer 1

D. 438g
Looking at the balanced reaction, for every 2 moles of Al(OH)3 consumed, 1 mole of Al2(SO4)3 is produced. So let's start by calculating the molar mass of the reactant and product.
Atomic weight aluminum = 26.981539
Atomic weight sulfur = 32.065
Atomic weight oxygen = 15.999
Atomic weight hydrogen = 1.00794

Molar mass Al(OH)3 = 26.981539 + (15.999 + 1.00794)*3 = 78.002359 g/mol
Molar mass Al2(SO4)3 = 26.981539 * 2 + (32.065 + 15.999*4)*3 = 342.146078 g/mol

Moles Al(OH)3 = 200.0 g / 78.002359 g/mol = 2.564025019 mol
Moles Al2(SO4)3 = 2.564025019 mol / 2 = 1.282012509 mol

Grams Al2(SO4)3 = 1.282012509 mol * 342.146078 g/mol = 438.635552 g
And the closest answer is "D. 438g"

Answer 2

the correct answer is 438g according to the following workings.

mole ratio of Aluminium hydroxide to aluminium sulphate= 2:1
number of moles in 200g of aluminium hydroxide = 200÷78
=2.564 moles
number of moles of Al2(SO4) produced by 200g of Al(OH)3=1.282
mass of Al2(SO4) produced = 342×1.282
= 438g