Answer to Question #184909 in Inorganic Chemistry for jurin junior

Question #184909
  1. A creation to convert ammonia into nitric acid involved following chemical equation:

4 NH3 (g) + 5 O2 (g)a 4 NO(g) + 6 H2O(g)

Assume that 1.50 g of NH3 react with 2.75 g O2, then;

a. determine the limiting reactant

b. calculate how many grams of NO and H2O will form in the reaction above

c. calculate how many grams of the excess reactant will remain after the reaction is complete.

d. if the yield obtained for NO is 1.80 grams and yield obtained for H2O is 1.50 grams, calculate the percent yield of NO and H2O in this reaction

(atomic mass of N=14; O=16; H=1)

Expert's answer

Mass of NH3 supplied = 1.50 g

Moles of NH3

Mass of O2 supplied

Moles of O2

(a) 1 mole of NH3 reacts with = 1.25 moles of O2

0.088 moles of NH3 reacts with = 0.11 moles of O2

Since, moles of O2 supplied = 0.085 moles

Therefore, O2 is limiting reactant.

(b) 5 moles of O2 forms 4 moles NO

1 mole of O2 forms moles NO

0.085 moles of O2 forms = 0.068 moles NO


Moles of H2O formed = 0.102 moles

NO formed =

H2O formed =

(c) 5 moles of O2 reacts 4 moles of NH3

1 mole of O2 reacts with moles of NH3

0.085 moles of O2 reacts with moles of NH3

Moles of NH3 unreacted = Moles Supplied Moles Reacted

Mass of NH3 unreacted =

(d) Yield obtained for NO = 1.80 g

NO formed theoretically = 2.04 g

Percentage Yield for NO =

Yield obtained for H2O = 1.50 g

H2O formed theoretically = 1.836 g

Percentage Yield for H2O =

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