Question

in the \before reaction\ box, there are 3 molecules of elemental nitrogen gas, and 3 molecules of elemental oxygen gas.
before reaction
legend
n₂
o₂
the contents of the box are allowed to react according to the unbalanced reaction below:
n₂(g) + o₂(g) → n₂o₃(g) (not balanced)
balance the reaction and then determine the theoretical yield of n₂o₃.
the theoretical yield is 4.5 molecules of n₂o₃.
the theoretical yield is 5 molecules of n₂o₃.
the theoretical yield is 6 molecules of n₂o₃.
the theoretical yield is 3 molecules of n₂o₃.
the theoretical yield is 2 molecules of n₂o₃.

Explanation:

Step1: Balance the chemical equation

The unbalanced reaction is \( \text{N}_2(g) + \text{O}_2(g)
ightarrow \text{N}_2\text{O}_3(g) \). To balance it, we need to adjust the coefficients. Let's balance the oxygen atoms first. On the right side, \( \text{N}_2\text{O}_3 \) has 3 O atoms, and on the left side, \( \text{O}_2 \) has 2 O atoms. The least common multiple of 2 and 3 is 6. So we put a coefficient of 3 in front of \( \text{O}_2 \) and a coefficient of 2 in front of \( \text{N}_2\text{O}_3 \). Then we balance the nitrogen atoms: \( 2\text{N}_2(g) + 3\text{O}_2(g)
ightarrow 2\text{N}_2\text{O}_3(g) \).

Step2: Determine the moles of reactants

We have 3 molecules of \( \text{N}_2 \) and 3 molecules of \( \text{O}_2 \). From the balanced equation, the mole ratio of \( \text{N}_2 \) to \( \text{O}_2 \) is 2:3. Let's check the limiting reactant. For \( \text{N}_2 \): if we have 3 molecules of \( \text{N}_2 \), according to the ratio 2:3, the required \( \text{O}_2 \) would be \( \frac{3}{2} \times 3 = 4.5 \) molecules, but we only have 3 molecules of \( \text{O}_2 \). For \( \text{O}_2 \): 3 molecules of \( \text{O}_2 \) would require \( \frac{2}{3} \times 3 = 2 \) molecules of \( \text{N}_2 \), and we have 3 molecules of \( \text{N}_2 \), so \( \text{O}_2 \) is the limiting reactant.

Step3: Calculate theoretical yield

From the balanced equation, 3 molecules of \( \text{O}_2 \) react with 2 molecules of \( \text{N}_2 \) to produce 2 molecules of \( \text{N}_2\text{O}_3 \)? Wait, no, let's recheck the mole ratio. The balanced equation is \( 2\text{N}_2 + 3\text{O}_2
ightarrow 2\text{N}_2\text{O}_3 \). So the ratio of \( \text{O}_2 \) to \( \text{N}_2\text{O}_3 \) is 3:2. So if we have 3 molecules of \( \text{O}_2 \), the moles of \( \text{N}_2\text{O}_3 \) produced would be \( \frac{2}{3} \times 3 = 2 \)? Wait, no, maybe I made a mistake. Wait, the balanced equation: 2 N₂ reacts with 3 O₂ to make 2 N₂O₃. So the mole ratio of O₂ to N₂O₃ is 3:2. So 3 molecules of O₂ would produce \( \frac{2}{3} \times 3 = 2 \) molecules of N₂O₃? Wait, no, let's do it again. Wait, the balanced equation is 2N₂ + 3O₂ → 2N₂O₃. So for 3 molecules of O₂, the number of N₂O₃ produced is (2/3)3 = 2? But wait, let's check the initial amounts. Wait, maybe I messed up the ratio. Wait, the balanced equation: 2 moles of N₂ react with 3 moles of O₂ to produce 2 moles of N₂O₃. So the ratio of N₂ to O₂ is 2:3. We have 3 N₂ and 3 O₂. Let's see how much each can produce. For N₂: 3 N₂ would need (3/2)3 = 4.5 O₂, but we only have 3 O₂. For O₂: 3 O₂ would need (2/3)*3 = 2 N₂. So O₂ is limiting. Then, from the balanced equation, 3 O₂ produce 2 N₂O₃? Wait, no, 3 O₂ (coefficient 3) produces 2 N₂O₃ (coefficient 2). So the ratio is 3 O₂ : 2 N₂O₃. So 3 O₂ would produce 2 N₂O₃? But the options have 2 as an option. Wait, let's check the options again. The options are 4.5, 5, 6, 3, 2. Wait, maybe I balanced the equation wrong. Wait, maybe the correct balanced equation is \( 2\text{N}_2 + 3\text{O}_2
ightarrow 2\text{N}_2\text{O}_3 \) is correct. Wait, let's count the atoms. Left: 4 N, 6 O. Right: 4 N, 6 O. Yes, that's balanced. Now, we have 3 N₂ (6 N atoms) and 3 O₂ (6 O atoms). From the balanced equation, 2 N₂ (4 N) reacts with 3 O₂ (6 O) to make 2 N₂O₃ (4 N, 6 O). So if we have 3 O₂ (6 O), we can use 2 N₂ (4 N) to make 2 N₂O₃. So the theoretical yield is 2 molecules of N₂O₃.

Answer:

The theoretical yield is 2 molecules of \( \text{N}_2\text{O}_3 \). (The option: "The theoretical yield is 2 molecules of \( \text{N}_2\text{O}_3 \).")