A500-g sample of al2(so4)3 is reacted with 450 g of 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 excess reagent are unreacted?

The chemical reaction is as follows:
Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3 
Explanation:
1mol of Al2(SO4)3 will react with 3mol Ca(OH)2 to produce 3mol CaSO4 and 2 mol Al(OH)3.
First we have to find the number of moles of Al2(SO4)3 :
Number of moles = Mass/ Molar mass
Mass of Al2(SO4)3 = 500g
Molar mass of Al2(SO4)3 = 342.15 g/mol
Number of moles = 500/342.15
Number of moles = 1.461 mol Al2(SO4)3
Multiplying the coeffecient of Ca(OH)2 with 1.461:
= 3*1.461 = 4.383 mol Ca(OH)2 
Now we have to find the number of moles of Ca(OH)2:
Mass of Ca(OH)2 = 450g
Molar mass of Ca(OH)2 = 74.09 g/mol 
Number of moles = 450/74.09
Number of moles =  6.074 mol Ca(OH)2
We need 4.383mol to react completely with the Al2(SO4)3, so the Ca(OH)2 is in excess, and the Al2(SO4)3 is the limiting reactant. 
Excess unreacted: 6.074-4.383 = 1.69mol Ca(OH)2 unreacted . 

The answer would be true