A 500.0 g sample of Al2(SO4)3 is reacted with 450.0 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? Al2(SO4)3(aq) + 3Ca(OH)2(aq) -> 2Al(OH)3(s) + 3CaSO4(s)

Answer : The limiting reagent in this reaction is, and number of moles of excess reagent is, 1.69 moles

Explanation : Given,

Mass of = 500.0 g

Mass of = 450.0 g

Molar mass of = 342.15 g/mol

Molar mass of = 74.1 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

From the balanced reaction we conclude that

As, 1 mole of react with 3 mole of

So, 1.461 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Number of moles of excess reagent = 6.073 - 4.383 = 1.69 moles

Therefore, the limiting reagent in this reaction is, and number of moles of excess reagent is, 1.69 moles

a corroding iron, if that's one of your choices