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)

Part A

Suppose that 4.8 mol NO2 and 1.0 mol H2O combine and react completely. Which reactant is in excess?

Part B

How many moles of the reactant in excess are present after the reaction has completed?

Answer 1

Here the equation is

3NO_{2 (g)} + H₂O _(l)   2HNO_{3 (l)} +     NO _(g)

we combine 4.8 mol  NO₂ with 1 mol of H₂O .

Part : A

According to the equation only 3 mol of NO₂ reacts with 1 mol of H₂O .

So, Only 3 mol of  NO₂ will be utilized and other amount is excess. We have taken NO₂ more than 3 mol .

So, here NO₂ is in excess.

Part : B

We know that as per the equation only 3 mol of NO₂ reacts with 1 mol of H₂O .the other amount will be excess.

Here we have taken 4.8 mole of NO_{2 .}

So, excess amount = 4.8 - 3

  = 1.8 mol

So, 1.8 mol of NO₂ will be  present after the reaction has completed.