A #4.50g# #NH_3# reacts with #4.50g# #O_2#. What is the limiting reactant and how much excess remains after the reaction?

Question

Naomi Aronson · Accepted Answer

The limiting reactant is oxygen. The amount of excess ammonia that remains after the reaction is 1.31 g.

The complete combustion of ammonia yields nitrogen gas #("N"_2")# and water. https://tutor.hix.ai Equilibrium Formula #"4NH"_3" + 3O"_2##rarr##"2N"_2" + 6H"_2"O"# We need to determine the molar masses of the reactants and one product. I chose #"N"_2"#. Molar Densities #"NH"_3":##"17.03052 g/mol"# #"O"_2":##"31.9988g/mol"# #"N"_2":##"28.0134 g/mol"# https://tutor.hix.ai Divide the given mass by the molar mass in moles of ammonia and water. #"NH"_3":##4.50"g NH"_3xx(1"mol NH"_3)/(17.03052"g NH"_3)="0.26423 mol NH"_3"# #"O"_2":##4.50"g O"_2xx(1"mol O"_2)/(31.9988"g O"_2)="0.14063 mol O"_2"# Limiting Reactant The limiting reactant is the reactant that yields the lesser amount of a product, in this case #"N"_2"#. To get nitrogen at the top of the balanced equation, multiply the moles of each reactant by the mole ratio. Next, multiply by the molar mass of the selected product, in this case nitrogen gas. #"mol reactant"xx("mol N"_2)/("mol reactant")xx"product molar mass"# #"NH"_3":##0.26423"mol NH"_3xx(2"mol N"_2)/(4"mol NH"_3)xx(28.0134"g N"_2)/(1"mol N"_2)="3.70 g N"_2"# #"O"_2":##0.14063"mol O"_2xx(2"mol N"_2)/(3"mol O"_2)xx(28.0134"g N"_2)/(1"mol N"_2)="2.63 g N"_2"# The limiting reactant is #"O"_2"# because it has the smallest yield of #"N"_2"#. The reactant in excess is #"NH"_3"#. Mass of Ammonia that Reacted #"mol O"_2xx("4mol NH"_3)/("3mol O"_2)xx"molar mass NH"_3# #0.14063"mol O"_2xx(4"mol NH"_3)/(3"mol O"_2)xx(17.0134"g NH"_3)/(1"mol NH"_3)="3.19 g NH"_3"# The Quantity of Extra Ammonia #4.50 "g NH"_3-"3.19g NH"_3"=1.31 g NH"_3"# The amount of #"NH"_3"# that remains is #"1.31 g"#.

Peyton Adams · Answer

undefined To determine the limiting reactant, first, calculate the number of moles of each reactant using their respective molar masses. Then, use the stoichiometry of the balanced chemical equation to see which reactant runs out first. The reactant that produces fewer moles of the product is the limiting reactant.

Next, calculate the amount of excess remaining by subtracting the amount of the limiting reactant consumed from the initial amount of the excess reactant.