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Ozone, o3 (48.0 g/mol), is a product in automobile exhaust by the reaction represented by the question NO2+O2 --> no + O3.rnWhat mass of ozone is predicted to form from the reaction of 2.0 g NO2 (46.0 g/mol) in a car's exhaust and excess oxygen?rn2.0 g O3rn1.1 g O3rn2.1 g O3rn1.8 g o3
Accepted Answer
Here's how to solve the problem:
1. Balanced Equation: The reaction is already balanced:
NO2 + O2 -→ NO + O3
2. Moles of NO2:
Convert grams of NO2 to moles using its molar mass (46.0 g/mol):
2.0 g NO2
(1 mol NO2 / 46.0 g NO2) = 0.043 mol NO2
3. Mole Ratio:
The balanced equation shows a 1:1 mole ratio between NO2 and O3. This means for every 1 mole of NO2 that reacts, 1 mole of O3 is produced.
4. Moles of O3:
Since the mole ratio is 1:1, 0.043 mol of NO2 will produce 0.043 mol of O3.
5. Mass of O3:
Convert moles of O3 to grams using its molar mass (48.0 g/mol):
0.043 mol O3
(48.0 g O3 / 1 mol O3) = 2.1 g O3
Therefore, the mass of ozone predicted to form is 2.1 g O3.
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