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`

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} NH_3=(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 =
`(1)/(2)* 1.09=0.545moles` of
`CaCN_2`

But as yield of reaction is 75.6 %, the amount of
`CaCN_2` needed is =
`(0.545)/(75.6)* 100=0.724`

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