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Ozone, O3, is produced in automobile exhaust by the reaction represented by the equation:
NO2 (g) + O2 (g) → NO (g) + O3 (g)
What mass of ozone is predicted to form from the reaction of 5.0 g NO2 in a car’s exhaust and excess oxygen?

Sagot :

Answer: 5.28 g of [tex]O_3[/tex] is produced.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} NO_2=\frac{5.0g}{46g/mol}=0.11moles[/tex]

The balanced chemical reaction is:

[tex]NO_2(g)+O_2(g)\rightarrow NO(g)+O_3(g)[/tex]

[tex]NO_2[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.

According to stoichiometry :

1 mole of [tex]NO_2[/tex] produces =  1 mole of [tex]O_3[/tex]

Thus 0.11 moles of [tex]NO_2[/tex] will produce=[tex]\frac{1}{1}\times 0.11=0.11moles[/tex]  of [tex]O_3[/tex]

 Mass of [tex]O_3=moles\times {\text {Molar mass}}=0.11moles\times 48g/mol=5.28g[/tex]

Thus 5.28 g of [tex]O_3[/tex] is produced.

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