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

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"

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