Answer:
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
