Respuesta :
Answer: 2.1 g mass of ozone([tex]O_{3}[/tex]) is predicted to form from the reaction of 2.0 g [tex]NO_{2}[/tex] in a car's exhaust and excess oxygen
Given information : Mass of [tex]NO_{2}[/tex] = 2.0 g and [tex]O_{2}[/tex] is in excess.
We need to calculate the mass of ozone ([tex]O_{3}[/tex])
Mass of ozone([tex]O_{3}[/tex]) is calculated with the help of mass of [tex]NO_{2}[/tex] using stoichiometry.
[tex]NO_{2} + O_{2}\rightarrow NO + O_{3}[/tex]
Step 1 : Convert grams of [tex]NO_{2}[/tex] to moles of [tex]NO_{2}[/tex].
[tex] Moles = \frac{Grams}{Molar mass}[/tex]
Molar mass of [tex]NO_{2}[/tex] = 46.0 g/mol
[tex]Moles = \frac{2.0g}{46.0\frac{g}{mol}}[/tex]
Moles of [tex]NO_{2}[/tex] = 0.043 mol
Step 2 : Find the moles of [tex]O_{3}[/tex] using moles of [tex]NO_{2}[/tex].
Moles of [tex]O_{3}[/tex] is calculated by using moles of [tex]NO_{2}[/tex] 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 [tex]NO_{2}[/tex] is 1 and coefficient of O3 is 1 , so mole ratio of [tex]O_{3}[/tex] to [tex]NO_{2}[/tex] is 1:1
Moles of [tex]O_{3}[/tex] = [tex](0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}[/tex]
Moles of [tex]O_{3}[/tex] = [tex](0.043)\times \frac{(1 mol O_{3})}{(1)}[/tex]
Moles of [tex]O_{3}[/tex] = 0.043 mol
Step 3 : Convert moles of [tex]O_{3}[/tex] to grams of [tex]O_{3}[/tex]
Grams = Moles X Molar mass
Molar mass of [tex]O_{3}[/tex] = 48.0 g/mol
Grams = [tex](0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})[/tex]
Grams = [tex](0.043)\times (\frac{48 g O_{3}}{1})[/tex]
Grams = 2.1 g [tex]O_{3}[/tex]
Note : The above three steps can also be done using a single step setup.
Grams of [tex]O_{3}[/tex] = [tex](2.0 gNO_{2})\times \frac{(1mol NO_{2})}{(46.0 g NO_{2})}\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}\times \frac{(48.0 g O_{3})}{(1 mol O_{3})}[/tex]
Grams of [tex]O_{3}[/tex] = [tex] (2.0 )\times \frac{(1)}{(46.0 )}\times \frac{(1)}{(1)}\times \frac{(48.0 g O_{3})}{(1)}[/tex]
Grams of [tex]O_{3}[/tex] = 2.1 grams
