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?

Question

Natalie Anton · Accepted Answer

See below

#3Mg(s)+N_2(g)=Mg_3N_2(s)# First, make sure the equation is balanced, which it is in this instance. Presuming that the limiting reactant is magnesium: 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#. Considering that the limiting reactant is nitrogen: 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.

Cooper Adams · Answer

undefined To determine the limiting reactant, you need to compare the amounts of each reactant present and use stoichiometry to calculate how much product each reactant can produce. Then, the reactant that produces the least amount of product is the limiting reactant. You can use the given balanced equation to calculate the moles of product formed from the given moles of each reactant. After comparing the moles of product formed, you can determine which reactant is limiting.