Question

A (6.12x10^1) .00 g sample of strontium reacts with an excess of nitrogen to form strontium nitride. What mass of strontium nitride forms?

Answer 1

Approx. `0.7*g`....

Explanation:

We need a stoichiometric reaction to represent the reaction...

`3Sr(s) + N_2(g) rarr Sr_3N_2(s)`

Alternatively....

`Sr(s) + 1/3N_2(g)rarr1/3Sr_3N_2`

This is in fact a redox equation....and dinitrogen is FORMALLY reduced to give two equiv of `N^(3-)`, i.e. nitride anion...

`"Moles of metal"=(0.612*g)/(87.621*g*mol^-1)=6.98xx10^-3*mol`

And given the stoichiometry, we can make one third an equiv of strontium nitride....i.e. a mass of ....

`1/3xx6.98xx10^-3*molxx290.87*g*mol^-1=0.677*g`

I think this nitriding reaction even occurs if you burn strontium in air....you get some of the oxide, as well as some of the nitride....