Calculate the number of oxygen atoms required to ccombine with 7 g of `N_(2)` to form `N_(2)O_(3)` when `80%` of `N_(2)` is converted to `N_(2)O_(3)`.

Options

A. `2.3xx10^(23)`
B. `3.6xx10^(23)`
C. `1.8xx10^(23)`
D. `5.4xx10^(21)`

Answer

To solve the problem step by step, we will follow the stoichiometric calculations based on the balanced chemical equation and the given data.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the formation of nitrogen trioxide (N₂O₃) from nitrogen (N₂) and oxygen (O₂) is:
\[ 2 \text{N}_2 + 3 \text{O}_2 \rightarrow 2 \text{N}_2\text{O}_3 \]

### Step 2: Calculate the number of moles of N₂ in 7 g
The molar mass of nitrogen (N₂) is 28 g/mol. Therefore, the number of moles of N₂ in 7 g can be calculated as:
\[ \text{Moles of } N_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{7 \text{ g}}{28 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 3: Determine the amount of N₂ converted to N₂O₃
Since 80% of N₂ is converted to N₂O₃, we calculate:
\[ \text{Moles of } N_2 \text{ converted} = 0.80 \times 0.25 \text{ moles} = 0.20 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of O₂ required
From the balanced equation, we see that 2 moles of N₂ require 3 moles of O₂. Therefore, for 0.20 moles of N₂:
\[ \text{Moles of } O_2 \text{ required} = \left( \frac{3 \text{ moles of } O_2}{2 \text{ moles of } N_2} \right) \times 0.20 \text{ moles of } N_2 = 0.30 \text{ moles of } O_2 \]

### Step 5: Calculate the number of oxygen atoms
To find the number of oxygen atoms, we first need to calculate the number of molecules in 0.30 moles of O₂ using Avogadro's number (6.022 × 10²³ molecules/mol):
\[ \text{Number of } O_2 \text{ molecules} = 0.30 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.8066 \times 10^{23} \text{ molecules} \]
Since each O₂ molecule contains 2 oxygen atoms, we multiply the number of O₂ molecules by 2:
\[ \text{Number of oxygen atoms} = 1.8066 \times 10^{23} \text{ molecules} \times 2 = 3.6132 \times 10^{23} \text{ atoms} \]

### Final Answer
The number of oxygen atoms required to combine with 7 g of N₂ to form N₂O₃ when 80% of N₂ is converted is approximately:
\[ \approx 1.8069 \times 10^{23} \text{ atoms} \]