posted by Ken on .
What mass of PbI2 can be formed from 75.0 mL of 0.65 M NaI by adding excess Pb(NO3)2?
Pb(NO3)2 + 2NaI ---> PbI + 2NaNO3
Step 1. Write and balance the equation. You have done that except the product on the right should be PbI2. That's just a typo, I know.
Step 2. Convert what you have been given ihnto mols. In this case it is M x L = mols.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (in this case mols NaI) to mol of what you want (in this case PbI2).
Step 4. Convert mols of the product you want (PbI2) to grams by using
grams = mols x molar mass.
Post your work if you get stuck.