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?

1 Answer
Jun 16, 2018

Answer:

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.