Question

Consider the reaction CaCN2 + 3 H2O → CaCO3 + 2 NH3 . This reaction has a 75.6% yield. How many moles of CaCN2 are needed to obtain 18.6 g of NH3?
a. 0.547 mol
b. 209 mol
c. 0.414 mol
d. 1.65 mol
e. 478 mol
f. 0.724 mol

Answer 1

Answer: Thus 0.724 mol of $CaCN_2$ are needed to obtain 18.6 g of $NH_3$

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} NH_3=\frac{18.6g}{17g/mol}=1.09moles$

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

According to stoichiometry :

2 moles of $NH_3$ are produced by = 1 mole of $CaCN_2$

Thus 1.09 moles of $NH_3$ will be produced by =$\frac{1}{2}\times 1.09=0.545moles$  of $CaCN_2$

But as yield of reaction is 75.6 %, the amount of $CaCN_2$ needed is =$\frac{0.545}{75.6}\times 100=0.724$

Thus 0.724 mol of $CaCN_2$ are needed to obtain 18.6 g of $NH_3$