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
See below
Explanation:
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