Question

Armonía is manufactured using the following reaction:
N2 + 3H2 ----- 2NH3

How much ammonia can be produced in the reaction of 2.04x10^-3 moles of nitrogen and 6.83x10^-3 moles of hydrogen?

Ammonia is manufactured using the following reaction: N2 + 3H2 - 2NH3. How much ammonia can be produced in the reaction of 2.04 x 10 moles of nitrogen and 6.83 x 10 moles of hydrogen?

Answer 1

SoC in first, calculate the limiting reagent Limiting Reagent > mole → Lowest limiting Reagent will be that whose male to stoichiometric ratio is lowest for Ny 7 m = 2:04x1003 - = 2:04x10-5 soc for the mole mobile - GosxlOS = 2:27470" LL+3H 6.83x10-3 _=2-27x10 IN +3H 2NH2 2017] Sic C S.C. i mole ole, ratio is lower for Ny, so N2 is limiting re a ratio is Sic N2+3H2 -> 2NH3 According to reaction imol of Ne produce a mal of NH3 • 200/103 mota produced Ammonia as a imol of Nz amol of NHS 2.048103 mol of a 2x2048103 mol of NHS

2:04X163 mol of N2 4:08x103 mol of NH3 As we know weight of one meal of NHg = 17 gram - so, 4.088103 not x17 grom Nits formed from 2004x10-3 mol of NL- mot - And, 6.936x102 gram of Ammonia formed from 20048103 mol of Ne so, Ammonia formared in mol = 4008 410² molt Ammonia formed in gram = 6.936 x102 gram