What is the mass of #"H"_2"# required to react with #"1.40 g N"_2"# in the reaction: #"N"_2 + "3H"_2"##rarr##"2NH"_3"#?

1 Answer
Nov 19, 2014

#"0.302 g H"_2"# is required to react with #"1.40 g N"_2"#.

Explanation:

You need to first convert the mass of nitrogen to moles using the molar mass of #"N"_2#, #"28.014g/mol"#. Then multiply the moles of nitrogen times the mole ratio between nitrogen and hydrogen in the balanced equation, such that the result is the number of moles of hydrogen. Then convert the moles of hydrogen to grams of hydrogen using the molar mass of #"H"_2#, #"2.016g/mol"#..

Start with a balanced equation.

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

Convert mass of #"N"_2# to moles.

#1.40color(red)cancel(color(black)("g N"_2))xx(1 "mol N"_2)/(28.014color(red)cancel(color(black)("g N"_2)))="0.049975 mol N"_2#

I am keeping a couple of extra digits to reduce rounding errors. I will round the final answer to three significant figures.

Calculate moles #"H"_2"#

Multiply the mol #"N"_2# by the mole ratio of #"3 mol H"_2/"1 mol N"_2# from the balanced equation.

#0.049975color(red)cancel(color(black)("mol N"_2))xx(3 "mol H"_2)/(1 color(red)cancel(color(black)("mol N"_2)))="0.14993 mol H"_2#

Convert mol of #"H"_2# to mass.

#0.14993color(red)cancel(color(black)("mol H"_2))xx(2.016"g H"_2)/(1 color(red)cancel(color(black)("mol H"_2)))="0.302 g H"_2# (rounded to three significant figures)