Question

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?

Answer 1

Answer:

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.