[asdasdasdasd]

I've tried the question myself and I've given my working out below. It is still some time before I can check and confirm with anyone whether I'm correct or not.

Here's the question,

"A chemist prepares Lead Chloride from Lead(II) Oxide and recorded the following masses,

Mass of PbO used in grams = 2.127g.
Mass of PbCl2 obtained = 2.523g.

The balanced equation shows that 1 mole of PbO gives 1 mole of PbCl2.

Calculate the % yield of the PbCl2."

This is my answer:
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n_(PbO) = mass/Molar Mass
          = 2.127/223.2
          = 9.530 x 10^-3 mol

Since 1 mole of PbO = 1 mole of PbCl2
Then, 1 mole of PbCl2 = 9.530 x 10^-3 mol

Therefore,

n_(PbCl2) = mass/Molar Mass

9.530 x 10^-3 mol =mass / 278.1
mass = 2.65g

% Yield = (actual yield/theoretical yield) * 100
           = (2.523 / 2.65) * 100

% Yield = 95.2%

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Thanks guys!

[Hunter2]

Sounds correct for me.