Question

hydrogen gas, H2, reacts with nitrogen gas,N2, to form ammonia gas , NH3 according to the equation...
3H2+N2 --->2NH3

1.how many grams of NH3 can be produced from 3.42 mol of N2 and excess H2
2. how many grams of H2 are needed to produce 14.93 g of NH3 ?
3. How many molecules (not moles) of NH3 are produced from 6.04*10^-4 g of H2 ?

Answer 1

1.

3.42 moles of N2 will produce NH3 = 2 x 3.42 moles of NH3

no. of moles of NH3 produced = 6.84 moles

weight of NH3 produced = 6.84 x 17 = 116.28 grams of NH3

2.

no. of moles of NH3 = 14.93/17 = 0.8782 moles

moles of H2 needed = (3/2) x 0.8782 = 1.31735 moles of H2

weight of H2 needed = 1.31735 x 2 = 2.6347 grams of H2

3.

no. of moles of H2 = (6.04*10^-4)/2 = 3.02*10^-4 moles

no. of moles of NH3 formed = (2/3) x 3.02*10^-4 moles = 2.013*10^-4 moles

no. of molecules formed = (2.013*10^-4) x (6.023*10^23) = 12.126 *10^19 molecules