Explanation:
<em>3Mg(s) + N2(g) = Mg3N2(s)</em>
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
<em>24.3g</em>, so the molecular weight is just <em>24.3g\mol</em>
2. Next we need the mole to mole ratio. As there are <em>3</em>
magnesiums for <em>1</em> magnesium nitride (shown by the coefficients), the
mole to mole ratio is<em> 1 mol Mg3N2\3 mol Mg.</em>
3. We need the amount of the substance, in grams. Since you have not
stated it in the question, I'll just do <em>10g</em> AS AN EXAMPLE. Note that
depending on the amount, the LIMITING REAGENT MAY DIFFER.
4. Finally, we need the molecular weight of <em>Mg3N2</em>, which we can easily
calculate to be around <em>100.9\mol.</em>
<em />
5. Putting this all together, we have<em> 10gMg⋅ (mol Mg\24.3gMg) </em>
<em> (1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)</em>
the units will cancel to leave <em>gMg3N2</em> (grams of magnesium nitride):
<em> 10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)</em>
<em> (100.9g Mg3N2\mol Mg3N2)</em>
<em />
Doing the calculation yields approximately 13.84g.
Assuming that nitrogen is the limiting reactant:
Similarly, following the above steps but with <em>10g</em> of nitrogen yields <em>36.04g</em>
In conclusion, as we produce less amount of <em>Mg3N2</em> when we assumed that <em>Mg</em> was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.
Note: This is in THIS CASE, where we have <em>10g</em> of both. The answer may vary depending on the amount of each substance.
