Question

A 2.50 kg sample of ammonia is mixed with 2.65 kg of oxygen.

A. Balance the equation below.

____ NH3(g) + ____ O2(g)  ____ NO(g) + ____ H2O(g)

B. Which is the limiting reagent?

C. How much excess reactant remains on completion of the reaction?

D. What is the theoretical yield of NO?

E. 1.45 kg of NO are produced what is the percent yield of NO?

Answer 1

NH3 moles = mass / molar mass = 2.5x1000 /17 =147.06

O2 moles = 2.65x1000 /32 = 82.8125

balanced eq is

A) 4 NH3 (g) + 5 O2 ---> 4NO (g) + 6H2O

B) as per reaction O2 moles needed ( 5/4) NH3 moles = ( 5/4) x 147 = 183.75 but we have 82.8125 moles

hence O2 is limiting reagent

C) NH3 moles reacted = (4/5) x O 2 moles = ( 4/5) x82.8125 = 66.25

NH3 moles left = 147.06-66.25 = 80.81

NH3 mass left = moles x molar mass = 80.81 x 17 = 1374 g = 1.374 kg

D) NO moles = ( 5/4) O2 moles = (5/4) x 82.8125 = 103.5156

NO mass = moles x molar mass = 103.5156 x 30.01 = 3105.5 g = 3.1055 kg

E) % yiled = 100 x actual yiled / tehoretical yiled = ( 100 x 1.45/3.1055) = 46.7 %