Question

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 1

Answer : The limiting reagent in this reaction is,
`Al_2(SO_4)_3` and number of moles of excess reagent is, 1.69 moles

Explanation : Given,

Mass of
`Al_2(SO_4)_3` = 500.0 g

Mass of
`Ca(OH)_2` = 450.0 g

Molar mass of
`Al_2(SO_4)_3` = 342.15 g/mol

Molar mass of
`Ca(OH)_2` = 74.1 g/mol

First we have to calculate the moles of
`Al_2(SO_4)_3` and
`Ca(OH)_2`.

`\text{Moles of }Al_2(SO_4)_3=\frac{\text{Given mass }Al_2(SO_4)_3}{\text{Molar mass }Al_2(SO_4)_3}`

`\text{Moles of }Al_2(SO_4)_3=(500.0g)/(342.15g/mol)=1.461mol`

and,

`\text{Moles of }Ca(OH)_2=\frac{\text{Given mass }Ca(OH)_2}{\text{Molar mass }Ca(OH)_2}`

`\text{Moles of }Ca(OH)_2=(450.0g)/(74.1g/mol)=6.073mol`

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

`Al_2(SO_4)_3(aq)+3Ca(OH)_2(aq)\rightarrow 2Al(OH)_3(s)+3CaSO_4(s)`

From the balanced reaction we conclude that

As, 1 mole of
`Al_2(SO_4)_3` react with 3 mole of
`Ca(OH)_2`

So, 1.461 moles of
`Al_2(SO_4)_3` react with
`1.461* 3=4.383` moles of
`Ca(OH)_2`

From this we conclude that,
`Ca(OH)_2` is an excess reagent because the given moles are greater than the required moles and
`Al_2(SO_4)_3` 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,
`Al_2(SO_4)_3` and number of moles of excess reagent is, 1.69 moles