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?

Lead nitrate Pb(NO3)2-Sodium iodide NaI

Sodium iodide moles are equal to 15//(22.99+126.9) = 15//149.89 = 0.100 moles.

To find the amount of sodium nitrate that can be formed, first, calculate the molar masses of lead (II) nitrate, sodium iodide, and sodium nitrate. Then, determine the limiting reactant using the stoichiometry of the reaction. Finally, use the limiting reactant to calculate the mass of sodium nitrate produced.

1. Molar mass of lead (II) nitrate (Pb(NO3)2): Pb: 207.2 g/mol N: 14.01 g/mol O: 16.00 g/mol Total: 207.2 + 2(14.01 + 3(16.00)) = 331.2 g/mol

2. Molar mass of sodium iodide (NaI): Na: 22.99 g/mol I: 126.90 g/mol Total: 22.99 + 126.90 = 149.89 g/mol

3. Molar mass of sodium nitrate (NaNO3): Na: 22.99 g/mol N: 14.01 g/mol O: 16.00 g/mol Total: 22.99 + 14.01 + 3(16.00) = 85.00 g/mol

4. Determine the limiting reactant: Pb(NO3)2 + 2NaI -> PbI2 + 2NaNO3 Using the molar masses and given masses, calculate the moles of each reactant: Moles of Pb(NO3)2 = 25.0 g / 331.2 g/mol Moles of NaI = 15.0 g / 149.89 g/mol The limiting reactant is the one that produces the least amount of product. Compare the moles of each reactant to the stoichiometry of the reaction to find the limiting reactant.

5. Calculate the mass of sodium nitrate produced: Since sodium nitrate is the product of the reaction involving the limiting reactant, use the moles of the limiting reactant and the molar mass of NaNO3 to find the mass of NaNO3 produced.

To find out how many grams of sodium nitrate can be formed, we first need to determine the limiting reactant in the reaction between lead(II) nitrate (Pb(NO3)2) and sodium iodide (NaI). This can be done by calculating the amount of sodium nitrate (NaNO3) produced by each reactant and comparing them. The reactant that produces the least amount of sodium nitrate will be the limiting reactant.

First, we need to write the balanced chemical equation for the reaction between lead(II) nitrate and sodium iodide:

Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3

From the balanced equation, we can see that 1 mole of lead(II) nitrate reacts with 2 moles of sodium iodide to produce 2 moles of sodium nitrate.

Next, we need to calculate the number of moles of each reactant:

For lead(II) nitrate: Number of moles = Mass / Molar mass Number of moles = 25.0 g / (207.2 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol)) Number of moles = 25.0 g / 331.22 g/mol Number of moles ≈ 0.0755 mol

For sodium iodide: Number of moles = Mass / Molar mass Number of moles = 15.0 g / (22.99 g/mol + 1.00 g/mol + 2(16.00 g/mol)) Number of moles = 15.0 g / 149.89 g/mol Number of moles ≈ 0.1001 mol

Now, let's determine the amount of sodium nitrate produced by each reactant:

For lead(II) nitrate: Number of moles of sodium nitrate produced = 0.0755 mol × (2 mol NaNO3 / 1 mol Pb(NO3)2) Number of moles of sodium nitrate produced ≈ 0.151 mol

For sodium iodide: Number of moles of sodium nitrate produced = 0.1001 mol × (2 mol NaNO3 / 2 mol NaI) Number of moles of sodium nitrate produced ≈ 0.1001 mol

Since the amount of sodium nitrate produced by lead(II) nitrate is greater than that produced by sodium iodide, sodium iodide is the limiting reactant.

Now, we can calculate the mass of sodium nitrate produced using the number of moles of sodium nitrate produced by sodium iodide:

Mass of sodium nitrate = Number of moles × Molar mass Mass of sodium nitrate = 0.1001 mol × (22.99 g/mol + 1.00 g/mol + 3(16.00 g/mol)) Mass of sodium nitrate ≈ 0.1001 mol × 85.99 g/mol Mass of sodium nitrate ≈ 8.60 g

Therefore, approximately 8.60 grams of sodium nitrate can be formed.