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.

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%