# Magnesium nitride may be prepared by the direct reaction of the elements as shown 3 mg(s)+n2(g)=Mg3N2(s). For each combination of the ff. combinations of reactants decide which is limiting reactant?

Jun 16, 2018

See below

#### Explanation:

$3 M g \left(s\right) + {N}_{2} \left(g\right) = M {g}_{3} {N}_{2} \left(s\right)$

First check that the equation is balanced. In this case, it is.

Assuming that magnesium is the limiting reactant:

1. First find the molecular weight using the Periodic Table.
We find that the atomic mass of magnesium is approximately $24.3 g$, so the molecular weight is just $\frac{24.3 g}{m o l}$
2. Next we need the mole to mole ratio. As there are $3$ magnesiums for $1$ magnesium nitride (shown by the coefficients), the mole to mole ratio is $\left(1 \text{mol "Mg_3N_2)/(3"mol } M g\right)$ .
3. We need the amount of the substance, in grams. Since you have not stated it in the question, I'll just do $10 g$ AS AN EXAMPLE. Note that depending on the amount, the LIMITING REAGENT MAY DIFFER.
4. Finally, we need the molecular weight of $M {g}_{3} {N}_{2}$, which we can easily calculate to be around $\frac{100.9 g}{m o l}$.
5. Putting this all together, we have 10"g"Mg*(("mol " Mg)/(24.3gMg))((1"mol "Mg_3N_2)/(3"mol " Mg))((100.9"g "Mg_3N_2)/("mol "Mg_3N_2))

the units will cancel to leave $\text{g} M {g}_{3} {N}_{2}$ (grams of magnesium nitride):

10cancel("g"Mg)*((cancel("mol " Mg))/(cancel(24.3gMg)))((cancel(1"mol "Mg_3N_2))/(cancel(3"mol " Mg)))((100.9"g "Mg_3N_2)/(cancel("mol "Mg_3N_2)))

Doing the calculation yields approximately $13.84 g$.

Assuming that nitrogen is the limiting reactant:

Similarly, following the above steps but with $10 g$ of nitrogen yields $36.04 g$

In conclusion, as we produce less amount of $M {g}_{3} {N}_{2}$ when we assumed that $M g$ was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

Note: This is in THIS CASE, where we have $10 g$ of both. The answer may vary depending on the amount of each substance.