posted by Dianna .
what is the percentage yield for
Sr(NO3)2 + Na2SO4 -> SrSO4 + 2 NaNO3.
mass of strontium nitrate is 0.788g
mass of sodium sulfate is 1.258g
mass of dry filter paper 0.759 g
mass of dried precipitate in filter paper 1.317g
This is a two part question; the first part is to determine the amount of yield for 100%. This is a limiting reagent(LR) problem because amounts are given for BOTH reactants.
mols Sr(NO3)2 = grams/molar mass
mols Na2SO4 = grams/molar mass
Using the coefficients in the balanced equation, convert mols each to mols SrSO4. It is likely you will obtain two different values which means one is wrong. In LR problems the correct value is ALWAYS the smaller value and the reagent responsible for that number is the LR. Theh g SrSO4 = mols x molar mass and this is the theoretical yield(TY) (at 100%).
Then %yield = (actual yield/TY)*100 = ?
Note: Actual yield is 1.317-0.759 = ?