Verified answer

1. the molecular weight of N2 is 14 X 2 = 28.0 g

b. you have 55.0 g of N2

c. therefore you have 55.0 g divided by 28.0 g/mol = 1.96 mol N2

2. the molecular weight of H2 gas is 2 g

b. therefore you have 27.5 mol of H2

The limiting reagent is N2 gas

3. The stoichiometry of the reaction is 1N2 : 3 H2 --> 2 NH3

you have 1.96 mol of the limiting reagent N2 (instead of the ration of 1 as above) - therefore multiply all molar quatities by 1.96 and you end up with 1.96 X 2 moles of NH3 which is 3.92 moles of NH3

The molecular weight of NH3 is 14 + 3(1) = 17.0 g

3.92 mol X 17 g/mol = 66.6 g of NH3 produced.

That is the beauty of molar quantities.

3h2 N2 2nh3

The limiting reactant is Nitrogen

55gram of Nitrogen gas is equivalent for = 55gm/14mg = 4 mole

the amount of ammonia gas produced is only 58gram

please give material balance