Question

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?

Answer 1

See below

Explanation:

`3Mg(s)+N_2(g)=Mg_3N_2(s)`

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 `24.3g`, so the molecular weight is just `(24.3g)/(mol)`
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 `(1"mol "Mg_3N_2)/(3"mol " Mg)` .
3. We need the amount of the substance, in grams. Since you have not stated it in the question, I'll just do `10g` AS AN EXAMPLE. Note that depending on the amount, the LIMITING REAGENT MAY DIFFER.
4. Finally, we need the molecular weight of `Mg_3N_2`, which we can easily calculate to be around `(100.9g)/(mol)`.
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 `"g"Mg_3N_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 `13.84g`.

Assuming that nitrogen is the limiting reactant:

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

In conclusion, as we produce less amount of `Mg_3N_2` when we assumed that `Mg` was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

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