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

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