Question

Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibrium mixture in a 5.00 L flask at 245°C contains 4.01 g of PCl5, 8.71 g of PCl3, and 2.87 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00 L flask at the same temperature? (Enter unrounded values.)

I'm pretty sure I did this problem right but my final answers were only 3 decimal places long unrounded. I carried my exact numbers through the entire problem

Answer 1

Phosphorus pentachloride decomposes according to this equation:

PCl_5(g) PCl_3(g) + Cl_2(g)

An equilibrium mixture in a 5.00 L flask at 245°C contains 4.01 g of PCl₅

8.71 g of PCl₃ and

2.87 g of Cl₂

Molarity of PCl₅ can be calculated as:

Molecular weight of PCl₅ = 208.24 g

1 mole of PCl₅ contains 208.24 g

Therefore, for 4.01g of PCl₅, the number of moles will be:

of PCl₅

Hence the molarity becomes:

Molarity of PCl₃ can be calculated as:

Molecular weight of PCl₃ = 137.33 g

1 mole of PCl₃ contains 137.33 g

Therefore, for 8.71 g of PCl₃, the number of moles will be:

of PCl₃

Hence the molarity becomes:

Molarity of Cl₂ can be calculated as:

Molecular weight of Cl₂ = 70.91 g

1 mole of Cl₂ contains 70.91 g

Therefore, for 2.87 g of Cl₂, the number of moles will be:

of Cl₂

Hence the molarity becomes:

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For PCl₅,

0.02 moles of PCl₅ is present in 5 L solution. Therefore, how many moles will be present in 2 L solution.

0.008 moles of PCl₅ will be 1.67 g of PCl₅.

For PCl₃,

0.06 moles of PCl₃ is present in 5 L solution. Therefore, how many moles will be present in 2 L solution.

0.024 moles of PCl₃ will be 3.29 g of PCl₃.

For Cl₂,

0.04 moles of Cl₂ is present in 5 L solution. Therefore, how many moles will be present in 2 L solution.

0.016 moles of Cl₂ will be 1.13 g of Cl₂.