Question

N2(g) + 3H2(g) → 2NH3(g)

5.00 g N2 is reacted with 5.00 g H2

How many mol NH3 can be produced in this reaction?

Answer 1

Number of moles of N2 = 5.00g / 28.0134 g/mol = 0.178 mole

Number of moles of H2 = 5.00g / 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

0.178 mole of N2 will require

= 0.178 mole of N2 *(3 mole of H2 / 1 mole of N2)

= 0.534 mole of H2

But we have 2.48 mole of H2 so H2 is excess reactant and N2 is limiting reactant

from the balanced equation we can say that

1 mole of N2 produces 2 mole of NH3 so

0.178 mole of N2 will produce

= 0.178 mole of N2 *(2 mole of NH3 / 1 mole of N2 )

= 0.356 mole of NH3

Therefore, the number of moles of NH3 produced will be 0.356 mole