Question

Ammonia (NH_3) is burned with pure oxygen to form nitric oxide (NO) and water Write the balanced chemical equation for this reaction If 60kg off ammonia are reacted with 110 kg of oxygen and the reaction proceeds to completion. Calculation the kgmols of all the resulting product compounds A mixture of ethyl alcohol and water at 20 degree c is prepared by mixing 5 L of 100 wt % alcohol (SG = 0.78934) with 4 L of pure water at 20 degree c (SG = 0.99823). Calculate the wt% alcohol in the resulting 100 proof mixture of alcohol and water. Calculate the volume (L) of the alcohol-water mixture if its SG = 0.9145?

Answer 1

(10)

(a) Balanced equation

4NH3 + 5O2 -------------> 4NO + 6H2O

(b) Moles of ammonia used = 60 * 1000 / 17 = 3529.4 mol

Moles of Oxygen used = 110 * 1000 / 32 = 3437.5 mol

According to balanced equation, 4 mole of ammonia requires 5 moles of oxygen.

Hence, oxygen is limiting reagent. (3437.5 < 3529.4)

From the balanced equation,

5 moles of oxygen can form 4 moles of NO and 6 moles of H2O

Then, 3437.5 moles of O2 can from 3437.5 * 4 / 5 moles of NO and 3437.5*6/5 moles of H2O

SO, moles of NO formed = 2750 mol

Mass of NO formed = 2750 * 30 /1000 = 82.5 kg.

Similarly, moles of H2O fromed = 4125 mol

Mass of H2O formed = 4125 * 18 / 1000 = 74.25 kg.