Answer to Question #184895 in Inorganic Chemistry for Toa

Question

1. A reaction to convert ammonia into nitric acid involves following chemical

equation:

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

Assume that 1.50 g of NH3 reacts 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

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

a.) NH_3 is the limiting reagent.

c.) Moles of Ammonia $= \dfrac{1.50}{17}$ = $0.088$ mol

Moles of Oxygen = $\dfrac{2.75}{32}$ = $0.086$ mol

Difference $= 0.088-0.086 = 2.0 \times 10^-{3}$ mol

Mass of excess reactant that remain $= 2.0 \times 10^{-3} \times 17$ = $0.034$ grams

Moles of NO $= \dfrac{1.80}{30} = 0.06$ moles

Moles of $H_2O = \dfrac{1.50}{33} = 0.045$ moles