Help me in solving Stoichiometry ??

Question

("Number") = subscript

A reaction mixture contains 21.4 grams of PCL(3) and 13.65grams of PbF(2).

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

3PbF(2) + 2PCl(3) --> 2PF(3) + 3PbCl(2)

How much of which reactant is left unchanged?

Answer 1

3PbF(2) + 2PCl(3) --> 2PF(3) + 3PbCl(2)

Molar mass PCl3 = 137.33 g/mol http://en.wikipedia.org/wiki/Phosphorus_...

Molar mass PbF2 = 245.20 g/mol http://en.wikipedia.org/wiki/Lead(II)_fl...

# moles = mass/molar mass

for PbF2

13.65/245.20=0.0556688=0.05567 moles PbF2

for PCl3

21.4/137.33=0.155829=0.1558 moles PCl3

so molar ratio for reaction from reaction equation 3 PbF2 : 2 PCl3

so 0.05567 moles PbF2 is the limiting reagent

So to find out how much PbCl2 can be formed look at the molar ratios again 3 PbF2 : 3 PbCl2 the molar ratio is 1:1.

so 0.05567 moles PbF2 would produce 0.05567 moles PbCl2

molar mass PbCl2 = 278.10 g/mol http://en.wikipedia.org/wiki/Lead(II)_ch...

mass = # moles * molar mass = 0.05567 moles PbCl2 * 278.10 g/mol = 15.482 g PbCl2

15.482 g PbCl2 produced by reaction

part B) From equation 3PbF(2) + 2PCl(3) --> 2PF(3) + 3PbCl(2)

we can see the molar ratio 3PbF2 : 2PCl3

we already calculated the moles of reactants we have

0.05567 moles PbF2

0.1558 moles PCl3

so to react the 0.05567 moles PbF2 we used 2/3 the number of moles PCl3. Or 0.05567 moles PbF2 *(2/3) = 0.03711 moles PbCl3

so we subtract what we used for the reaction from what we started with

0.1558 moles PCl3-0.03711 moles PCl3=0.11869 moles PCl3

0.11869 moles PCl3  left after reaction

or

Molar mass PCl3 = 137.33 g/mol http://en.wikipedia.org/wiki/Phosphorus_...

0.11869 moles PCl3  * 137.33 g/mol = 16.2996 or 16.3 g PCl3 left unreated

Answer 2

You've got to first determine the limiting reagent:

13.65 g PbF2 x (1 mol / 245.20 g PbF2) x (2 mol PCl / 3 mol PbF2) x ( 137.25 g / mol PCl3) = 5.09 g PCl3 needed for 13.65 g PbF2

That means that PbF2 is your limiting reagent (you've got more than 4 times the necessary mass of PCl3).  (Also, you'll have 16.3 g PCl3 left over).

Now, how much PbCl2 will be made from 13.65 g PbF2?

13.65 g PbF2 x (1 mol / 245.20 g PbF2) x ( 3 mol PbCl2 / 3 mol PbF2) x ( 278.1 g / 1 mol PbCl2) = 15.48 g PbCl2 produced.