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?

2.1 g mass of ozone() is predicted to form from the reaction of 2.0 g in a car's exhaust and excess oxygen

Given information : Mass of = 2.0 g and is in excess.

We need to calculate the mass of ozone ()

Mass of ozone() is calculated with the help of mass of using stoichiometry.

Step 1 : Convert grams of to moles of .

Molar mass of = 46.0 g/mol

Moles of = 0.043 mol

Step 2 : Find the moles of using moles of .

Moles of is calculated by using moles of 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 is 1 and coefficient of O3 is 1 , so mole ratio of to is 1:1

Moles of =

Moles of =

Moles of = 0.043 mol

Step 3 : Convert moles of to grams of

Grams = Moles X Molar mass

Molar mass of = 48.0 g/mol

Grams =

Grams =

Grams = 2.1 g

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

Grams of =

Grams of =

Grams of = 2.1 grams

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