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

1 Answer
Nov 23, 2015

Answer:

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

Explanation:

The complete combustion of ammonia yields nitrogen gas #("N"_2")# and water. https://en.wikipedia.org/wiki/Ammonia

Balanced Equation

#"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 Masses

#"NH"_3":##"17.03052 g/mol"#
#"O"_2":##"31.9988g/mol"#
#"N"_2":##"28.0134 g/mol"#
http://pubchem.ncbi.nlm.nih.gov/

Moles of Ammonia and Water
Divide given mass by molar mass.

#"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"#.

For each reactant, multiply moles of reactant times the mole ratio from the balanced equation so that nitrogen is on top, then multiply times the molar mass of the chosen 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 Amount of Excess 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"#.