How to solve this help me !!?

Question

how to solve this guys please..

A reaction mixture contains 21.4 grams of PCL3 and 13.65grams of PbF2.

What mass of PBCl2 can be obtained from the following reaction?

3PbF2 + 2PCl3 --> 2PF3 + 3PbCl2

How much of which reactant is left unchanged?

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

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as an edit:

Were you to prefer to work this problem emphasizing the ratio of moles to moles,

the ratio of moles of PbF2 : moles PCl3 is as follows:

3 : 2 = 0.0556 : x

x = moles PCl3 needed =0.0371

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& the ratio between moles  PbF2 :  moles PbCl2 is 2 : 2

Moles PbCl2 = 2 x 0.0556 / 2 =0.0556

And the Mass PbCl2 = 0.0556 mol x 278.10 g/mol =15.46 g

which when I rounded off earlier to 3 sig fgis, gave you the same answer:

15.5 g of PbCl2 can be produced

Answer 2

Moles PCl3 = 21.4 g / 137.33 g/mol =0.156

Moles PbF2 = 13.65 g / 245.30 g/mol =0.0556

3 : 2 = x : 0.0556

x = moles PCl3 needed =0.0834

Moles PCl3 in excess = 0.156 - 0.0834 =0.0726

Mass PCl3 in excess = 0.0726 mol x 137.33 g/mol = 9.97 g

the ratio between PF2 and PbCl2 is 2 : 3

Moles PbCl2 = 3 x 0.0556 / 2 =0.0834

Mass PbCl2 = 0.0834 mol x 278.10 g/mol =23.2 g