Question

Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --> 3no(g) + o3(g). what mass of ozone is predicted to form from the reaction of 2.0 g no2 in a car's exhaust and excess oxygen?

Answer 1

Answer: 2.1 g mass of ozone(
`O_(3)`) is predicted to form from the reaction of 2.0 g
`NO_(2)` in a car's exhaust and excess oxygen

Given information : Mass of
`NO_(2)` = 2.0 g and
`O_(2)` is in excess.

We need to calculate the mass of ozone (
`O_(3)`)

Mass of ozone(
`O_(3)`) is calculated with the help of mass of
`NO_(2)` using stoichiometry.

`NO_(2) + O_(2)\rightarrow NO + O_(3)`

Step 1 : Convert grams of
`NO_(2)` to moles of
`NO_(2)`.

`Moles = (Grams)/(Molar mass)`

Molar mass of
`NO_(2)` = 46.0 g/mol

`Moles = (2.0g)/(46.0(g)/(mol))`

Moles of
`NO_(2)` = 0.043 mol

Step 2 : Find the moles of
`O_(3)` using moles of
`NO_(2)`.

Moles of
`O_(3)` is calculated by using moles of
`NO_(2)` with the help of mole ratio.

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of
`NO_(2)` is 1 and coefficient of O3 is 1 , so mole ratio of
`O_(3)` to
`NO_(2)` is 1:1

Moles of
`O_(3)` =
`(0.043 mol NO_(2))* ((1 mol O_(3)))/((1 mol NO_(2)))`

Moles of
`O_(3)` =
`(0.043)* ((1 mol O_(3)))/((1))`

Moles of
`O_(3)` = 0.043 mol

Step 3 : Convert moles of
`O_(3)` to grams of
`O_(3)`

Grams = Moles X Molar mass

Molar mass of
`O_(3)` = 48.0 g/mol

Grams =
`(0.043 mol O_(3))* ((48 g O_(3))/(1 mol O_(3)))`

Grams =
`(0.043)* ((48 g O_(3))/(1))`

Grams = 2.1 g
`O_(3)`

Note : The above three steps can also be done using a single step setup.

Grams of
`O_(3)` =
`(2.0 gNO_(2))* ((1mol NO_(2)))/((46.0 g NO_(2)))* ((1 mol O_(3)))/((1 mol NO_(2)))* ((48.0 g O_(3)))/((1 mol O_(3)))`

Grams of
`O_(3)` =
`(2.0 )* ((1))/((46.0 ))* ((1))/((1))* ((48.0 g O_(3)))/((1))`

Grams of
`O_(3)` = 2.1 grams