Question

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation 3H2(g)+N2(g)→2NH3 How many grams of H2 are needed to produce 10.70 g of NH3? How many molecules (not moles) of NH3 are produced from 3.21×10−4 g of H2?

Answer 1

number of moles of NH3 = 10.70g / 17.031 g/mol = 0.628 mole

from the balanced equation we can say that

2 mole of NH3 is produced by 3 mole of H2 so

0.628 mole of NH3 will be produced by 0.942 mole of H2

1 mole of H2 = 2.016 g

0.942 mole of H2 = 1.90 g

Therefore, the mass of H2 produced would be 1.90g

number of moles of H2 = 3.21*10^-4 g / 2.016 g/mol = 1.59*10^-4 mole

from the balanced equation we can say that

3 mole of H2 produces 2 mole of NH3 so

1.59*10^-4 mole of H2 will produce 1.06*10^-4 mole of NH3

1 mole of NH3 = 6.023*10^23 molecules

1.06*10^-4 mole of NH3 = 6.38*10^19 molecules

Therefore, the number of molecules of NH3 = 6.38*10^19 molecules