How many grams of nitrogen are necessary to produce 15.00 grams of ammonia?

1 Answer

The required mass of nitrogen is 12.34 g.

Explanation:

The balanced equation is:

0.5"N"_2 + 1.5"H"_2 rarr "NH"_3 color(white)(..................................)(1)

The molecular mass of ammonia, "NH"_3, is equal to 14.01 g + 3(1.008 g), or 17.03 g.

"Moles of ammonia" = "mass in grams"/"molecular mass" ="15.00 g"/"17.03 g/mol" = 15.00/17.03 " mol"

Therefore, according to equation (1),

"Moles of nitrogen" = 0.5(15.00/17.03) " mol" color(white)(............) (2)

The number of moles of nitrogen is calculated by the following equation:

"Moles of N"_2 = ("mass of nitrogen")/ ("molar mass of nitrogen")

Thus,

"Mass of nitrogen" = "moles of N"_2*"molar mass of N"_2 color(white)(...) (3)

Therefore, from equations (2) and (3), we have:

"Mass of nitrogen" = 0.5(15.00/17.03) " mol" xx "28.02 g/mol" = "12.34 g"

With kind regards
Dr. Mamdouh Younes
Hamilton, Ontario, Canada