Answer : The excess reactant is, ![H_2O](https://tex.z-dn.net/?f=H_2O)
The leftover amount of excess reagent is, 7.2 grams.
Solution : Given,
Mass of
= 105.0 g
Mass of
= 78.0 g
Molar mass of
= 80.11 g/mole
Molar mass of
= 18 g/mole
Molar mass of
= 100.09 g/mole
First we have to calculate the moles of
and
.
![\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DCaCN_2%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DCaCN_2%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DCaCN_2%7D%3D%5Cfrac%7B105.0g%7D%7B80.11g%2Fmole%7D%3D1.31moles)
![\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DH_2O%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DH_2O%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DH_2O%7D%3D%5Cfrac%7B78.0g%7D%7B18g%2Fmole%7D%3D4.33moles)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
![CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3](https://tex.z-dn.net/?f=CaCN_2%2B3H_2O%5Crightarrow%20CaCO_3%2B2NH_3)
From the balanced reaction we conclude that
As, 1 mole of
react with 3 mole of ![H_2O](https://tex.z-dn.net/?f=H_2O)
So, 1.31 moles of
react with
moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
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.
Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles
Now we have to calculate the mass of excess reactant.
![\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20excess%20reactant%7D%3D%5Ctext%7B%20Moles%20of%20excess%20reactant%7D%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20excess%20reactant%7D%28H_2O%29)
![\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20excess%20reactant%7D%3D%280.4moles%29%5Ctimes%20%2818g%2Fmole%29%3D7.2g)
Thus, the leftover amount of excess reagent is, 7.2 grams.