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

May 7, 2018

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....

${\underbrace{M {g}_{3} {N}_{2} \left(s\right) + 6 {H}_{2} O \left(l\right)}}_{\text{209 g" rarr underbrace(3Mg(OH)_2(aq) + 2NH_3(aq))_"209 g}}$

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 ...........
$\text{moles of magnesium salt} = \frac{100 \cdot g}{100.95 \cdot g \cdot m o {l}^{-} 1} = 0.991 \cdot m o l$

$\text{moles of water} = \frac{75.0 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 4.16 \cdot m o l$

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.