1. Write a balanced equation for lead(II) nitrate with sodium iodide to form sod

Question

1. Write a balanced equation for lead(II) nitrate with sodium iodide to form sodium nitrate and lead(II) iodide.
2. If I start with 25.0 grams of lead(II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
3. What is the limiting reagent in problem 2?
4. How much of the non limiting reagent will be left over from the reaction in problem 2? 1. Write a balanced equation for lead(II) nitrate with sodium iodide to form sodium nitrate and lead(II) iodide.
2. If I start with 25.0 grams of lead(II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
3. What is the limiting reagent in problem 2?
4. How much of the non limiting reagent will be left over from the reaction in problem 2?
2. If I start with 25.0 grams of lead(II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
3. What is the limiting reagent in problem 2?
4. How much of the non limiting reagent will be left over from the reaction in problem 2? 2. If I start with 25.0 grams of lead(II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
3. What is the limiting reagent in problem 2?
4. How much of the non limiting reagent will be left over from the reaction in problem 2?

Explanation / Answer

1. Write a balanced equation for lead(II) nitrate with sodium iodide to form sodium nitrate and lead(II) iodide.

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

2. If I start with 25.0 grams of lead(II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

Mol wt of Pb(NO3)2 is 331.2 g/ mol

Mol wt of Sodium iodide is 150 g/ mol

as from above equation 331.2 g of lead nitrate reacts with 300 g of sodium iodide ( 2 moles) to give 170 g of sodium nitrate ( 2moles)

so 15 g of NaI will react with 16.56 g of lead nitrate to give 8.5 g of sodium nitrate

3. What is the limiting reagent in problem 2?

Sodium iodide is limiting reagent

4. How much of the non limiting reagent will be left over from the reaction in problem 2

left ove rof excess reagent ( lead nitrate) is 25-16.56g = 8.44 g

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