Limiting Reagent question...?

Question

A 500 g sample of Al2 (SO4)3 is reacted with 450 g Ca(OH)2. A total of 596 g of CaSO4 is produced. What is the limiting reagent in this reaction, and how many moles of reagent are unreacted?

Al2 (SO4)3 + 3Ca (OH)2 --> 2Al(OH)3 + 3CaSO4

can anyone show me how to do this problem? i'm completely lost!

Answer 1

Step 1: Find the molar mass for each compound where a mass is given, and then find the # of moles.

Molar masses:
Al2 (SO4)3's molar mass = 2 * Al + 3 * S + 12 * O
= 2 * 27 + 3 * 32 + 12 * 16 = 54 + 96 + 192 = 342 g/mol
Ca(OH)2 = Ca + 2O + 2H = 40 + 32 + 2 = 74 g/mol
CaSO4 = Ca + S + 4O = 40 + 32 + 64 = 136 g/mol

# of moles:
Al2 (SO4)3's = 500g / 342 g/mol = 1.462 moles
Ca(OH)2 = 450g / 74 g/mol = 6.081 moles
CaSO4 = 596g / 136 g/mol = 4.382 moles

Step 2: Find the ratios between each reagent and the # of moles of CaSO4, as determined by the reaction equation. Then, find the ratios as determined by the actual masses. Whichever ratios match - that's your limiting reagent.

Reaction equation:
Ratio of Al2 (SO4)3 : CaSO4 = 1:3
Ratio of Ca(OH)2 : CaSO4 = 1:1

Molar ratio:
Ratio of Al2 (SO4)3 : CaSO4 = 1.462 : 4.382 = 1:3
Ratio of Ca(OH)2 : CaSO4 = 6.081 : 4.382 = 1.39:1

The ratios match for Al2 (SO4)3, so that must be the limiting reagent.