Explanation:
3Mg(s) + N2(g) = Mg3N2(s)
First check that the equation is balanced. In this case, it is.
Assuming that magnesium is the limiting reactant:
- 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 Mg3N2\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 Mg3N2, which we can easily
calculate to be around 100.9\mol.
5. Putting this all together, we have 10gMg⋅ (mol Mg\24.3gMg)
(1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)
the units will cancel to leave gMg3N2 (grams of magnesium nitride):
10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)
(100.9g Mg3N2\mol Mg3N2)
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 Mg3N2 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.