Answer: Thus 0.724 mol of [tex]CaCN_2[/tex] are needed to obtain 18.6 g of [tex]NH_3[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} NH_3=\frac{18.6g}{17g/mol}=1.09moles[/tex]
[tex]CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3[/tex]
According to stoichiometry :
2 moles of [tex]NH_3[/tex] are produced by = 1 mole of [tex]CaCN_2[/tex]
Thus 1.09 moles of [tex]NH_3[/tex] will be produced by =[tex]\frac{1}{2}\times 1.09=0.545moles[/tex] of [tex]CaCN_2[/tex]
But as yield of reaction is 75.6 %, the amount of [tex]CaCN_2[/tex] needed is =[tex]\frac{0.545}{75.6}\times 100=0.724[/tex]
Thus 0.724 mol of [tex]CaCN_2[/tex] are needed to obtain 18.6 g of [tex]NH_3[/tex]
