Consider the reaction of 2g and H2g 1a How many grams of NH3

Consider the reaction of 2(g) and H2(g)

1a. How many grams of NH3 can be produced (theoretically from the reaction of 5.0 g of N2
and 5.0 g H2?
b. What is the limiting reagent?
c. If 5.80 g of NH3 are actually formed, what is the percent yield?

Solution

Balanced equation is

N2 + 3 H2 -----> 2 NH3

number of moles of N2 = 5.0 g / 28.0 g/mol = 0.179 mole

number of moles of H2 = 5.0 g / 2.016 g/mol = 2.48 mole

from the balanced equation we can say that

1 mole of N2 requires 3 mole of H2 so

0.179 mole of N2 will require

= 0.179 mole of N2 *(3 mole of H2 / 1 mole of N2)

= 0.537 mole of H2

But we have 2.48 mole of H2 which is in excess so

N2 is the limiting reactant

from the balanced equation we can say that

1 mole of N2 produces 2 mole of NH3 so

0.179 mole of N2 will produce

= 0.179 mole of N2 *(2 mole of NH3 / 1 mole of N2)

= 0.358 mole of NH3

mass of 1 mole of NH3 = 17.031 g

so the mass of 0.358 mole of NH3 = 6.10 g

Therefore, theoretical yield of NH3 = 6.10 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (5.80 / 6.10)*100

percent yield = 95.1 %

Therefore, percent yield = 95.1 %