Question

Ammonia reacts with O2 to form either NO or NO2 according to the unbalanced equations shown below: NH3 + O2 → NO + H2O NH3 + O2 → NO2 + H2O In an experiment, 10 mol NH3 and 15.1 mol O2 are intially placed in a closed flask. After the reaction all the NH3 has reacted and 2.4 mol O2 remain. Calculate the number of moles of NO in the flask after reaction. Round your answer to two significant figures.

Answer 1

The number of moles of NO would be 10.0 moles.

First, the equation of the reaction leading to the production of NO needs to be balanced. Thus:

$4NH_3 + 5O2 ---> 4NO + 6H_2O$

From the equation, 4 moles of NH3 produces 4 moles of NO. This means that the ratio of NH3 input into the reaction and NO produced is 1:1.

Also, since all the NH3 was used up and O2 was in excess, the limiting reactant is NH3.

Thus, if 10 mol of NH3 was initially placed in the closed flask and all of it was used up in reaction, 10 mol of NO would also be produced.

Hence, the number of moles of NO in the flask after reaction would be 10.0 moles to two significant figures.

More on calculating number of moles in reactions can be found here: https://brainly.com/question/9734304