Question

When 100 g `Mg_3N_2` reacts with 75.0 g `H_2O`, what is the limiting reactant?

Answer 1

The salt REQUIRES SIX EQUIV of water....and the water is the limiting reactant....

Explanation:

As always, our priority is to write a stoichiometrically balanced equation to represent the reaction....

`underbrace(Mg_3N_2(s) + 6H_2O(l))_"209 g" rarr underbrace(3Mg(OH)_2(aq) + 2NH_3(aq))_"209 g"`

I subscripted the masses of reactants and product to establish that the equation was stoichiometrically kosher. Garbage out EQUALS garbage in, as is absolutely required.

And so ...........
`"moles of magnesium salt"=(100*g)/(100.95*g*mol^-1)=0.991*mol`

`"moles of water"=(75.0*g)/(18.01*g*mol^-1)=4.16*mol`

And so here UNUSUALLY the water is in stoichiometric deficiency. And it is unusual in that in these circumstances it is hard to exclude water from a reaction setup. The nitride may be incompletely reduced to give say hydrazine or some other product.