Given:
- Mass of N2 = 7g
- Percentage of N2 converted into products = 80%
- Formula of N2O3

To Find: Number of oxygen atoms required to combine with 7g of N2 to form N2O3

Solution:

Step 1: Write the balanced chemical equation for the reaction
N2 + xO2 → N2O3
where 'x' represents the number of oxygen atoms required.

Step 2: Calculate the amount of N2 used in the reaction
The molar mass of N2 = 28 g/mol
Number of moles of N2 = Mass of N2/Molar mass of N2 = 7/28 = 0.25 mol

Step 3: Calculate the amount of N2O3 formed
Since 80% of N2 is converted into products, the number of moles of N2O3 formed can be calculated as follows:
Number of moles of N2O3 = 80/100 × 0.25 = 0.2 mol

Step 4: Calculate the number of oxygen atoms required
From the balanced chemical equation, we can see that one molecule of N2O3 contains 3 oxygen atoms. Therefore, the number of oxygen atoms required to form 0.2 mol of N2O3 can be calculated as follows:
Number of oxygen atoms = 0.2 × 3 × Avogadro's number
where Avogadro's number = 6.022 × 10^23 mol^-1

Number of oxygen atoms = 3.6 × 10^23

Answer: Therefore, the number of oxygen atoms required to combine with 7g of N2 to form N2O3 is 3.6 × 10^23.