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)

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04-04-2020
Chemistry

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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)

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Alleei Alleei · Answer 1 · 2020-04-06T14:13:49+02:00

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

Explanation : Given,

Mass of [tex]Al_2(SO_4)_3[/tex] = 500.0 g

Mass of [tex]Ca(OH)_2[/tex] = 450.0 g

Molar mass of [tex]Al_2(SO_4)_3[/tex] = 342.15 g/mol

Molar mass of [tex]Ca(OH)_2[/tex] = 74.1 g/mol

First we have to calculate the moles of [tex]Al_2(SO_4)_3[/tex] and [tex]Ca(OH)_2[/tex].

[tex]\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}[/tex]

[tex]\text{Moles of }Al_2(SO_4)_3=\frac{500.0g}{342.15g/mol}=1.461mol[/tex]

and,

[tex]\text{Moles of }Ca(OH)_2=\frac{\text{Given mass }Ca(OH)_2}{\text{Molar mass }Ca(OH)_2}[/tex]

[tex]\text{Moles of }Ca(OH)_2=\frac{450.0g}{74.1g/mol}=6.073mol[/tex]

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

[tex]Al_2(SO_4)_3(aq)+3Ca(OH)_2(aq)\rightarrow 2Al(OH)_3(s)+3CaSO_4(s)[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]Al_2(SO_4)_3[/tex] react with 3 mole of [tex]Ca(OH)_2[/tex]

So, 1.461 moles of [tex]Al_2(SO_4)_3[/tex] react with [tex]1.461\times 3=4.383[/tex] moles of [tex]Ca(OH)_2[/tex]

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