Question

If `30.9*g` calcium nitride are obtained from `65.5*g` of metal, what is the percentage yield of the reaction?

Answer 1

Approx. `40%`...

Explanation:

We need (i) a stoichiometric equation....

`3Ca(s) + N_2(g) stackrel(Delta)rarrCa_3N_2(s)`

And (ii), we need equivalent quantities of metal and nitride..

`"Moles of metal"=(65.5*g)/(40.1*g*mol^-1)=1.63*mol`

The dinitrogen reagent is ASSUMED to be in excess.....

..we need equivalent quantities of metal and nitride..

`"Moles of calcium nitride"=(30.9*g)/(148.25*g*mol^-1)=0.208*mol`

And given the stoichiometry....`"percentage yield"` is given by...

`"Moles of calcium nitride"/(1/3xx"moles of calcium")xx100%=(0.208*mol)/(1/3xx1.63*mol)xx100%=??%`

When we heat an active metal with dinitrogen, the process of nitrogen fixation is called `"nitriding"`...it is not a cost effective method of producing ammonia.