The feed to an ammonia synthesis reactor contains 25 mole% nitrogen and the balance hydrogen. the fow rate of the stream is 3000 kg/h. calculate the rate of flow of nitrogen into the reactor in kg/h. (suggestion: first calculate the average molecular weight of the mixture.)

The rate of flow of nitrogen into the reactor is 2,470.588 kg/h.

Explanation:

Mole percentage of nitrogen gas  = 25 mole%

Mole percentage of hydrogen  gas  = 75 mole%

Average molecular weight of the mixture:

Rate of flow of the stream = 3000 kg/h

Mass of stream in 1 hour = 3000 kg = 3,000,000 g

Moles of stream :

Moles of nitrogen gas in 352,941.17 moles of stream be x

x = 88,235.294 mol

Mass of 8,823,529.4 mole of nitrogen gas:

The rate of flow of nitrogen into the reactor is 2,470.588 kg/h.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

these nuts