Balance the below equation: NH_3+ 〖O〗_2 → NO+ H_2 O
How many grams of NO can be produced from 12 g of NH3 and 12 g of O2?

What is the limiting reactant? What is the excess reactant?

How much excess reactant remains when the reaction is over?

O₂ is the limiting reactant

0.406 moles of ammonia remains after the reaction goes complete.

Explanation:

Balanced reaction is:

4NH₃  + 5O₂  →  4NO  +  6H₂O

Let's determine the moles of each reactant:

12 g . 1mol / 17g = 0.706 moles of ammonia.

12 g . 1mol / 32g = 0.375 moles of oxygen.

4 moles of ammonia react to 5 moles of oxygen

Then, 0.706 moles of ammonia may react to (0.706 . 5) /4 = 0.882 moles

We only have 0.375 moles of oxygen and there are needed 0.882. O₂ is the limiting reactant. Definetely ammonia is in excess.

5 moles of oxygen react to 4 moles of ammonia

Our 0.375 moles may react to (0.375 . 4) / 5 = 0.3 moles

We have 0.706 moles of NH₃ and we only need 0.3

After the reaction goes complete (0.706 - 0.3) = 0.406 moles of ammonia still remains.

b

explanation:

mitosis

c) should be the answer.