P5-F2 + Ar --> ArF3 P5a- How many moles of F2 (Fluorine) are needed to produce 9 mole of ArF3 (Argon trifluoride)?
P5b- How many moles of Argon (Ar) are consumed in a reaction that consumes 5 moles of F2 (fluorine)?

I need the answers for the p5 part please help

5a. 13.5 moles of F₂

5b. 3.33 moles of Ar.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3F₂ + 2Ar —> 2ArF₃

From the balanced equation above,

3 moles of F₂ reacted with 2 moles of Ar to produce 2 moles of ArF₃.

5a. Determination of the number of mole of fluorine, F₂, needed to produce 9 moles of ArF₃.

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of ArF₃.

Therefore, Xmol of F₂ will react to produce 9 moles of ArF₃ i.e

Xmol of F₂ = (3 × 9)/2

Xmol of F₂ = 13.5 moles

Thus, 13.5 moles of F₂ is needed.

5b. Determination of the number of mole of Ar needed to react with 5 moles F₂.

From the balanced equation above,

3 moles of F₂ reacted with 2 moles of Ar.

Therefore, 5 moles of F₂ will react with = (5 × 2)/3 = 3.33 moles of Ar.

Thus, 3.33 moles of Ar was consumed.

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

1. they vary with a higher number because boiling is way hotter than melting