Answer:
B - The amount of OF2(g) produced is halved.
Explanation:
Because its right.
When instead of 2 moles of F₂, 1 mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.
Let's consider the balanced reaction that occurs when F₂ reacts with NaOH.
2 F₂(g) + 2 NaOH(aq) → OF₂(g) + 2 NaF(aq) + H₂O(l)
First, let's see the moles of OF₂ and NaF obtained from 2 moles of F₂, using the molar ratios derived from the balanced chemical equation.
[tex]2 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 1 mol OF_2\\\\2 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 2 mol NaF[/tex]
Now, let's compare with the moles of OF₂ and NaF obtained from 1 mol of F₂, again using the same molar ratios derived from the balanced chemical equation.
[tex]1 mol F_2 \times \frac{1 molOF_2}{2mol F_2} = 0.5 mol OF_2\\\\1 mol F_2 \times \frac{2 molNaF}{2mol F_2} = 1 mol NaF[/tex]
As we can see, since we have half the amount of F₂, we obtain half the amount of the products. Then, the only right option is: The amount of OF₂(g) produced is halved.
When instead of 2 moles of F₂, 1 mole of F₂ reacts with excess NaOH, the amount of OF₂(g) produced is halved.
You can learn more about molar ratios here: https://brainly.com/question/9408497
