The following reaction was performed:
Pb(NO3)2 + NaI → PbI2 + NaNO3
25.2 grams of lead nitrate is reacted with 15.5 grams of sodium iodide to form 6.2 grams of sodium nitrate.
Step 1 - Balance the equation (put a 1 in the blank if there is no coefficient).
1 Pb(NO3)2 + 2 NaI → 1 PbI2 + 2 NaNO3
Step 2 - Calculate the theoretical yield of NaNO3
For Pb(NO3)2 → Theoretical Yield NaNO3 13 g - round to the nearest tenth.
For NaI → Theoretical Yield NaNO3 9 g - round to the nearest tenth.
Step 3 - Identifying the Limiting Reagent (which reactant had the lowest TY?)
Limiting Reagent = NaI
Step 4 - Percent yield of NaNO3 (calculate the % yield of the LR)
Percent yield % - round to the nearest tenth.
