A reaction was run where lead nitrate and potassium iodide were mixed. The resulting precipitate only gave a 78.1% yield. If the actual yield was 25.0grams, how many grams of lead nitrate were used?

Question

This is what I have so far:

Potassium Iodide= KI = 166.0028g/mol
lead nitrate= Pb(NO3)2 =331.2 g/mol

I do not see how I can obtain the grams of nitrate from the information that is given can someone please help me?

Answer 1

The reaction is:
Pb(NO3)2 + 2KI ---> PbI2(s) + 2KNO3(aq)
1) Solve: 0.781x = 25.0g (x = theoretical yield of PbI2)
2) Convert the theoretical yield of PbI2 to moles.
3) Moles of Pb(NO3)2 = moles of PbI2 = _______?
4) Convert moles of Pb(NO3)2 to grams (Multiply # moles by the formula mass).

Answer 2

2KI + Pb(NO3)2 ==>2KNO3 + PbI2

You know the actual yield was 25.0 grams. Since it was 78.1% yield, then
%yield = [(actual yield/theoreticalyield)]*(100 = 78.1.
Substitute 25.0 for actual yield and calculate theoretical yield. Convert to moles PbI2,
then use the coefficients in the balanced equation to convert moles PbI2 to moles Pb(NO3)2, then convert that to grams. Post your work if you get stuck.

Answer 3

When I see those questions my brain explodes but once I work it thorugh it makes so much sense to me:)

Thanks for the help