# How many moles of magnesium oxide are produced by the reaction of 3.82 g of magnesium nitride with 7.73 g of water? Mg_3N_2 + 3H_2O -> 2NH_3 3MgO

Aug 2, 2017

We assess the rxn....

$M {g}_{3} {N}_{2} \left(s\right) + 6 {H}_{2} O \left(l\right) \rightarrow 2 N {H}_{3} \left(a q\right) + 3 M g {\left(O H\right)}_{2} \left(a q\right)$

#### Explanation:

Is it balanced? Don't trust my arithmetic.

This tells us UNEQUIVOCALLY that $100.95 \cdot g$ $\text{magnesium nitride}$ reacts with $108.06 \cdot g$ water to give stoichiometric $\text{ammonia}$, and $\text{magnesium hydroxide}$.

We have molar quantities of........

$\text{moles of}$ $M {g}_{3} {N}_{2} = \frac{3.82 \cdot g}{100.95 \cdot g \cdot m o {l}^{-} 1} = 0.0378 \cdot m o l$

$\text{moles of}$ $O {H}_{2} = \frac{7.73 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 0.429 \cdot m o l$

Water is thus present in stoichiometric excess. Do you agree?

And thus we get stoichiometric $\text{magnesium hydroxide}$, i.e. $0.0378 \cdot m o l \times 3 \cdot m o l \times 58.32 \cdot g \cdot m o {l}^{-} 1 = 6.61 \cdot g$.