Gaseous ammonia burns in oxygen in accordance with: 4NH_3 (g) + 3O_2 (g) rightar
ID: 1021461 • Letter: G
Question
Gaseous ammonia burns in oxygen in accordance with: 4NH_3 (g) + 3O_2 (g) rightarrow2N_2 (g) + 6H_2O (l) In one experiment, equal masses of ammonia and oxygen gas were reacted, and 64.5 g of water is obtained, at a yield of 84.3%. Which reactant was limiting? How many grams of the excess reactant remains? (Note there were no side reactions; the reactions simply did not run to completion) How many grams of Nitrogen gas were produced in this experiment? What was the grand total of mass in the reaction mixture after the reaction? (i.e., mass of remaining reactants and products)Explanation / Answer
a)limiting reagent is that reactant ,which completely consumed in reaction.
let mass of NH3 and o2 is x.
84.3% of x is = 64.5g
> 84.3x / 100 = 64.5
x=64.5 x 100 / 84.3
x = 76.51g
number of mole of NH3 = 76.51 /17.03
=4.49
number of mole of o2 =76.51 / 16
=4.78
4mole of NH3 use 3 mole of O2
1 mole of NH3 use 3/4(0.75) mole of O2
NH3 will completely consumed.hens it is limiting.
b).
1 mole of NH3 use 3/4(0.75) mole of O2
4.49 mole of nh3 need 0.75 x 4.49 =3.367 mole of o2
remain mole of O2 is = 4.78 - 3.367
= 1.413 mol
mass of o2 remain in g = 1.413 x 16
=22.6g
c).
4mole of NH3 gives 2 mole of N2
1mole of NH3 gives 0.5 mole of N2
4.49 mole of Nh3 produce 0.5 x 4.49=2.245 mol
mass of N2 = 2.245 x 28
= 62.86g
d) mass of reactant remain + mass of product (H2O and N2)
22.6 + 64.5 +62.86 =149.96g
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