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?

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?

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).

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.

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

To find the grams of lead nitrate used, we can use the concept of stoichiometry.

First, we need to determine the molar ratio between lead nitrate (Pb(NO3)2) and the precipitate formed (PbI2) in the reaction. From the chemical equation:

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

we can see that 1 mole of lead nitrate reacts with 1 mole of lead iodide.

Given that the actual yield of lead iodide is 25.0 grams, we can calculate the molar amount of lead iodide:

Molar mass of PbI2 = 331.2 g/mol
Moles of PbI2 = Mass of PbI2 / Molar mass of PbI2 = 25.0 g / 331.2 g/mol

Next, using the stoichiometry ratio, we know that 1 mole of PbI2 is formed from 1 mole of Pb(NO3)2. Hence, the moles of Pb(NO3)2 used is also 25.0 g / 331.2 g/mol.

Now, we can find the grams of lead nitrate used:

Mass of Pb(NO3)2 = Moles of Pb(NO3)2 × Molar mass of Pb(NO3)2
= (25.0 g / 331.2 g/mol) × 331.2 g/mol

Simplifying, we get:

Mass of Pb(NO3)2 ≈ 25.0 g

Therefore, approximately 25.0 grams of lead nitrate were used in the reaction.