A challange question on chemistry.?

Question

A student weighs out 5.00 g of potassium iodide to react with an excess of lead II nitrate souliton. When the reaction is finished, weighs out the precipitate to a mass of 5.50 g. Calculate the student's percentage error. Help me with this one. Will be greatly appreciated.

Answer 1

Please note that from the balanced equation,

2 KI + Pb(NO3)2 -------- 2 K(NO3)2 + PbI2
2 moles of Potassium Iodide yields only one mole of the precipitate PbI2.

Thus, 5.00 g KI/Molar Mass =166 = 0.0301 moles of KI to produce 0.0151 moles of Lead Iodide.

But the Molar Mass of Lead Iodide is 209 + 127 + 127 =463g/mole

The theoretical amount in grams of the precipitate is
= 0.0151 x 463 = 6.99g

But you only got 5.50 g, so your error is a negative one, at

(6.99 -5.50 )/6.99 then multiply by 100 for the % error.

Or, = to 21.3 % error

Answer 2

I'll just give u a hint here. Coz right now i dont have a calculator and a periodic table. I suggest for all askers, it would be better if they give all the atomic weights or formula weights of the elements/compounds to ease in aswering calculation problems.

First write the balanced equation for the reaction.

Pb(NO3)2 + 2KI ---- PbI2 + 2KNO3

From the solubility rule, which will precipitate? It is the PbI. Not KNO3 since nitrates are soluble.

Given the grams of KI we can calculate the grams of PbI2 that will precipitate from the balanced equation.

mass PbI2 = [grams KI/FW KI]*[1mole PbI2/2moles KI]* FW PbI2

that mass of PbI2 u calculated is the theoretical yield. to calculate the percentage error:

%error = [theoretical yield - actual yield]*100/ theoretical yield

note the value in the numerator must be the absolute value

Hope this helps u

Answer 3

2KI + Pb(NO3)2 -------- 2K(NO3)2 + PbI2
MW=166
0.03 mole 0.03 mole

0.03 * [39 + 2(14+48)] = 0.03*163=4,89g
(5,50-4,89/4,89)*100=12,47%