# chemistry

posted by .

Stoichiometry

0.083grams of lead nitrate, Pb(NO3)2, in solutions mixed with 0.300g of sodium iodide in solution. This results in the formation of a yellow precipitate of lead iodide. Calculate the weight of the precipitate.

• chemistry -

Pb(NO3)2 + 2NaI ==> PbI2 + 2NaNO3

Convert 0.083 g Pb(NO3)2 to moles. #moles = grams/molar mass
Convert moles Pb(NO3)2 to moles PbI2 using the coefficients in the balanced equation which I wrote.
Convert moles PbI2 to grams. grams = moles x molar mass.

• chemistry -

the 0.300g is not needed?

• chemistry -

Yes, it IS needed but I overlooked it. However, there are 0.002 moles of NaI which is more than enough to react with all of the Pb(NO3)2 (0.00025 moles); therefore, the Pb(NO3)2 is the limiting reagent and the problem is worked exactly as I showed it. That is, the PbI2 formed will be the result of the 0.00025 moles (0.083 g) Pb(NO3)2.

## Respond to this Question

 First Name School Subject Your Answer

## Similar Questions

1. ### Chemistry

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?
2. ### Chemistry

For each of the following, write a net ionic equation (modeled above). Wherever there is a precipitate, state the name of the precipitate (e.g “lead iodide” from above). Pop onto the net and see if you can find a description of …
3. ### chemistry

if 5liters of hydrogen iodide gas is bubbled through 2 liters of 0.0275 M lead 2 nitrate solution at STP, how many grams of lead 2 iodide will precipitate out?
4. ### Chemistry

I have 5.0 ml of lead II NITRATE and mix it with 20.0 ml of Sodium Iodide. what is the formula of the precipitate
5. ### chemistry

why different masses of precipitate were obtained when various volume of solution lead nitrate and potassium iodide mixed
6. ### Chemistry

If 32.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.545 g precipitate, what is the molarity of lead(II) ion in the original solution?
7. ### chemistry

Determine the mass of the precipitate, lead iodide that is formed when 150 ml of 0.0500M pb(no3)2 is reacted with excess potassium iodide. if 3.3 of the precipitate is experimentally recovered, determine the percent yeild.
8. ### Chemistry

A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide solution. A bright yellow Precipitate is of lead(ii) iodide is formed. Calculate the mass of lead (ii) iodide that should be formed. …
9. ### Chemistry

We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. If we have 500. mL of 0.632 M lead(II) nitrate solution and add excess potassium iodide, how much solid lead(II) iodide …
10. ### Chemistry

I'm sorry here is another I do not understand. How do you find out exactly how much of a solution you would need to completely precipitate another compound in the reaction?

More Similar Questions