Answer to Question #184909 in Inorganic Chemistry for jurin junior

Question

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)

Answer 1

$4 NH_3 (g) + 5 O_2 (g)\longrightarrow 4 NO(g) + 6 H_2O(g)$

Mass of NH₃ supplied = 1.50 g

Moles of NH₃ $=\dfrac{1.50}{17}=0.088\space mol$

Mass of O₂ supplied $={2.75}\space g$

Moles of O₂ $=\dfrac{2.75}{32}=0.085\space mol$

(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 $=\dfrac{4}{5}$ moles NO

0.085 moles of O₂ forms = 0.068 moles NO

Similarly,

Moles of H₂O formed = 0.102 moles

NO formed = $0.068\times30=2.04\space g$

H₂O formed = $0.102\times18=1.836\space g$

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

1 mole of O₂ reacts with $=\dfrac{4}{5}$ moles of NH₃

0.085 moles of O₂ reacts with $=0.068$ moles of NH₃

Moles of NH₃ unreacted = Moles Supplied $-$Moles Reacted$=0.02\space mol$

Mass of NH₃ unreacted = $0.02\times17=0.34\space g$

(d) Yield obtained for NO = 1.80 g

NO formed theoretically = 2.04 g

Percentage Yield for NO = $\dfrac{1.80}{2.04}\times100=88.23\%$

Yield obtained for H₂O = 1.50 g

H₂O formed theoretically = 1.836 g

Percentage Yield for H₂O = $\dfrac{1.50}{1.836}\times 100=81.699\%$