If 6.5 mol NO2 react with 1.0 mol H20, how many moles of the excess reactant

remains when the reaction is complete? 3NO2 + H2O + NO + 2HNO3

1.5 mol H20

2.5 mol H20

O 3.5 mol NO2

O 1.0 mol NO2

3.5 moles of NO2.

Explanation:

The balanced equation for the reaction is given below:

3NO2 + H2O - > NO + 2HNO3

Next, we shall determine the excess reactant. This is illustrated below:

From the balanced equation above,

3 moles of NO2 reacted with 1 moles H2O.

Therefore, 6.5 moles of NO2 will react with = (6.5 x 1) / 3 = 2.17 moles of H2O.

From the above calculations, we can see that it will take a higher amount i. e 2.17 moles than what was given i. e 1 mole of H2O to react completely with 6.5 moles of NO2.

Therefore, NO2 is the excess reactant and H2O is the limiting reactant.

Next, we shall determine the number of mole of the excess reactant that reacted in the reaction. This is illustrated below:

From the balanced equation above, we can see that 3 moles of NO2 reacted with 1 mole of H2O.

Finally, we shall determine the leftover mole of the excess reactant after the reaction. This is illustrated below:

Given mole of NO2 = 6.5 moles

Reacted mole of NO2 = 3 moles

Leftover mole of NO2 = ... ?

Leftover mole = Given mole - Reacted mole

Leftover mole of NO2 = 6.5 - 3

Leftover mole of NO2 = 3.5 moles.

Therefore, 3.5 moles of NO2 remained after the reaction.