If 25 grams of NaI is mixed with excess amount Pb (NO3) 2, what would be the product?

If 25 grams of NaI is mixed with an excess amount of Pb (NO3) 2 we get 38.45 grams of PbI2 and 14.18 grams of NaNO3

Explanation:

Step 1: Data given

Mass of NaI = 25.00 grams

Pb (NO3) 2 is in excess

Step 2: The balanced equation

2NaI + Pb (NO3) 2 → PbI2 + 2NaNO3

Step 3: Calculate moles of NaI

Moles NaI = mass NaI / molar mass NaI

Moles NaI = 25.00 / 149.89 g/mol

Moles NaI = 0.1668 moles

Step 4: Calculate moles of PbI2 and NaNO3

For 2 moles of NaI we need 1 mol of Pb (NO3) 2 to produce 1 mol of PbI2 and 2 moles of NaNO3

For 0.1668 moles of of NaI we will have 0.1668/2 = 0.0834 moles of PbI2 and 0.1668 moles of NaNO3

Step 5: Calculate mass of the products

Mass PbI2 = 0.0834 * 461.01 g/mol

Mass PbI2 = 38.45 grams

Mass NaNO3 = 0.1668 mol * 84.99 g/mol

Mass NaNO3 = 14.18 grams

If 25 grams of NaI is mixed with an excess amount of Pb (NO3) 2 we get 38.45 grams of PbI2 and 14.18 grams of NaNO3