Consider the following equilibrium:
2NO(g) + 3F₂(g)
2NOF₃(g)
Suppose 0.20 mol of NO and 0.30 mol of F₂ are added to a 5.0-L container.If x mol of NOF₃ is present at equilibrium,how many moles of fluorine are present at equilibrium?

Question

Consider the following equilibrium:
2NO(g) + 3F₂(g)

Consider the following equilibrium: 2NO(g) + 3F<sub>2</sub>(g) 2NOF<sub>3</sub>(g) Suppose 0.20 mol of NO and 0.30 mol of F<sub>2</sub> are added to a 5.0-L container.If x mol of NOF<sub>3</sub> is present at equilibrium,how many moles of fluorine are present at equilibrium? A) 0.30 - x B) 0.30 - 2x C) 0.06 - 0.20x D) 0.30 - x E) 0.20 - x

2NOF₃(g)
Suppose 0.20 mol of NO and 0.30 mol of F₂ are added to a 5.0-L container.If x mol of NOF₃ is present at equilibrium,how many moles of fluorine are present at equilibrium?

Quizplus · Accepted Answer

From the balanced equation, we can see that for every 2 moles of NO reacted, 2 moles of NOF₃ are formed, and for every 3 moles of F₂ reacted, 2 moles of NOF₃ are formed. Therefore, we know that the amount of NOF₃ formed will be limited by the reactant that runs out first, which in this case is NO since we have an excess of F₂. Using an ICE table, we can set up the equation: 2NO(g) + 3F₂(g) ↔ 2NOF₃(g) I: 0.20 mol 0.30 mol 0 C: -2x -3x +2x E: 0.20-2x 0.30-3x x At equilibrium, we have 0.20 - 2x mol of NO remaining, 0.30 - 3x mol of F₂ remaining, and x mol of NOF₃ formed. From the balanced equation, we can see that there are 2 moles of fluorine in every mole of NOF₃, therefore the total amount of fluorine present at equilibrium is 2x mol. Thus, the answer is D) 0.30 - 11ea7a3a_9f1a_d92d_a82d_93f632eac194_TB2288_11 x.

Consider the Following Equilibrium