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?
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)#
- 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)#.
- 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.
- Finally, we need the molecular weight of
#Mg_3N_2#, which we can easily calculate to be around #(100.9g)/(mol)#.
- 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
Doing the calculation yields approximately
Assuming that nitrogen is the limiting reactant:
Similarly, following the above steps but with
In conclusion, as we produce less amount of
Note: This is in THIS CASE, where we have