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
.
![\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}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DAl_2%28SO_4%29_3%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DAl_2%28SO_4%29_3%7D%7B%5Ctext%7BMolar%20mass%20%7DAl_2%28SO_4%29_3%7D)
![\text{Moles of }Al_2(SO_4)_3=\frac{500.0g}{342.15g/mol}=1.461mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DAl_2%28SO_4%29_3%3D%5Cfrac%7B500.0g%7D%7B342.15g%2Fmol%7D%3D1.461mol)
and,
![\text{Moles of }Ca(OH)_2=\frac{\text{Given mass }Ca(OH)_2}{\text{Molar mass }Ca(OH)_2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCa%28OH%29_2%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DCa%28OH%29_2%7D%7B%5Ctext%7BMolar%20mass%20%7DCa%28OH%29_2%7D)
![\text{Moles of }Ca(OH)_2=\frac{450.0g}{74.1g/mol}=6.073mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCa%28OH%29_2%3D%5Cfrac%7B450.0g%7D%7B74.1g%2Fmol%7D%3D6.073mol)
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)](https://tex.z-dn.net/?f=Al_2%28SO_4%29_3%28aq%29%2B3Ca%28OH%29_2%28aq%29%5Crightarrow%202Al%28OH%29_3%28s%29%2B3CaSO_4%28s%29)
From the balanced reaction we conclude that
As, 1 mole of
react with 3 mole of ![Ca(OH)_2](https://tex.z-dn.net/?f=Ca%28OH%29_2)
So, 1.461 moles of
react with
moles of ![Ca(OH)_2](https://tex.z-dn.net/?f=Ca%28OH%29_2)
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