A reaction mixture contains 21.4 grams of PCl3 and 13.65 grams of PbF2. What mass of PbCl2 can be obtained ?

Question

A reaction mixture contains 21.4 grams of PCl3 and 13.65 grams of PbF2. What mass of PbCl2 can be obtained ?

Answer 1

3PbF2 + 2PCl3 --> 2PF3 + 3PbCl2 react as:

3PbF2 : 3 moles @ 245.20 g/mol = 735.6 grams PbF2

2PCl3 : 2 moles @ 137.33 g/mol = 274.66 grams PCl3

3PbCl2: 3 moles @ 278.10 g/mol = 834.3 grams PbCl2

3PbF2 / 2PCl3 react at a ratio of 735.6 grams / 274.66 grams = 2.7

however they added: 13.65 g PbF2 / 21.4 g PCl3 = 0.6

they have seriously limited the PbF2, it is the "limiting reagent"

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What mass of PbCl2 can be obtained from the following reaction?

13.65 g PbF2 @ 834.3 g PbCl2 / 735.6 g PbF2 = 15.48

your first answer is: 15.5 g PbCl2 can be produced

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How much of which reactant is left unchanged?

we will 1st find out how much PCl3 reacted with the PbF2:

13.65 g PbF2 @ 274.66 grams PCl3 / 735.6 g PbF2 = 5.10 grams of PCl3 reacted

now how much PCl3 is left over:

21.4 g PCl3 - 5.10 grams PCl3 reacted = 16.4 g

your last answer is : 16.4 grams of PCl3 is left over