What mass of ammonia can be produced by the reaction of #"2.16 mol N"_2"# and excess #"H"_2"#?

1 Answer
Sep 26, 2016

Answer:

The reaction will produce #"73.6 g NH"_3"#.

Explanation:

Balanced Equation

#"3H"_2"+N"_2"##rarr##"2NH"_3"#

Molar mass #"NH"_3"#: #"17.03052 g/mol"# https://pubchem.ncbi.nlm.nih.gov/compound/222#section=Top

Mole ratios between #"NH"_3"# and #"N"_2"#.

#(1"mol N"_2)/(2"mol NH"_3)# and #(2"mol NH"_3)/(1"mol N"_2")#

Multiply the given moles of nitrogen gas by the mole ratio that has #"NH"_3"# in the numerator. This gives the moles of #"NH"_3"# produced in this reaction. Then multiply by the molar mass of #"NH"_3"# to get the mass of #"NH"_3# produced in this reaction.

#2.16cancel"mol N"_2xx(2cancel"mol NH"_3)/(1cancel"mol N"_2)xx(17.03052"g NH"_3)/(1cancel"mol NH"_3)="73.6 g NH"_3"#