Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3no2(g)+h2o(l)→2hno3(l)+no(g) suppose that 11 mol no2 and 3 mol h2o combine and react completely. how many moles of the reactant in excess are present after the reaction has completed?

2 moles of the reactant are present in excess after the reaction has been completed

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide

The equation of reaction is:

3NO₂(g)  +   H₂O(l) >>>>> 2HNO₃(l) +  NO(g)

1 mole of H₂O reacts with 3 moles of NO₂

3 moles of H₂O will react with 9 moles of NO₂

Therefore, if 3 moles of H₂O reacts with 11 moles of NO₂( instead of 9 moles):

Number of moles of excess reactant = 11 moles - 9 moles

Number of moles of excess reactant = 2 moles

2 moles of the reactant are present in excess after the reaction has been completed

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2 mol NO2

Explanation:

                                      3NO2(g)+H2O(l)→2HNO3(l)+NO(g)

from reaction                3 mol        1 mol

given                           11 mol          3 mol

for 3 mol NO2     1 mol H2O

for x mol NO2     3 mol H2O

3:x = 1:3

x = 3 *3/1 = 9 mol NO2

So, for 3 mol H2O are needed only 9 mol NO2.

But we have 11 mol NO2. So, NO2 is in excess, and

11 mol NO2 - 9 mol NO2 = 2 mol NO2 will be left after reaction.

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