How many grams of #H_2# are needed to produce 11.27 g of #NH_3#?

#1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#

#11.27# #g# of ammonia represents #(11.27*g)/(17.03*g*mol^-1)# #=# #??# #mol#.

Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?

Start with a balanced equation.

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

Since the question concerns the mass of hydrogen gas needed, the assumption is that nitrogen gas is in excess.

The molar masses of #"H"_2"# and #"NH"_3"# are needed.

#"H"_2":# #"2.01588 g/mol"#
http://pubchem.ncbi.nlm.nih.gov/compound/783

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

We will be making the following conversions: mass #"NH"_3"##rarr##"mol NH"_3"##rarr##"mol H"_2"##rarr##"mass H"_2"#

1. Determine moles of #"NH"_3"# by dividing the given mass by its molar mass.

2. Determine moles of #"H"_2"# by multiplying the mole ratio from the balanced equation between #"H"_2"# and #"NH"_3"# with moles #"H"_2"# in the numerator. #(3"mol H"_2)/(2"mol NH"_3)#

3. Determine mass of #"H"_2"# by multiplying moles #"H"_2"# by its molar mass.

#11.27"g NH"_3xx(1"mol NH"_3)/(17.03052"g NH"_3)xx(3"mol H"_2)/(2"mol NH"_3)xx(2.01588"g H"_2)/(1"mol H"_2)="2.001 g H"_2"# (rounded to four significant figures)

2.001 g hydrogen gas is needed to produce 11.27 g ammonia if nitrogen gas is present in excess.