posted by Kevin .
Show the calculation for the limiting reagent when 0.853g of sodium sulfate (95% pure) react with 1.707g of barium hydroxide octahydrate (87% pure) to produce a precipitate of barium sulfate.
I've determined the equation is Na2SO4+Ba(OH)2*8H2O=BaSO4+2NaOH
However, that is unbalanced and I'm not sure how to balance the equation with the hydrated compound.
What would the first steps be to balance each side?
You have it balanced except for the 8H2O. Just add 8H2O on the end.
Na2SO4+Ba(OH)2*8H2O=BaSO4+2NaOH + 8H2O
g Na2SO4 = 0.853 x 0.95 = ?
g Ba(OH)2*8H2O = 1.707 x 0.87 = ?
Convert to mols.
mols Na2SO4 = grams/molar mass = ?
mols Ba(OH)2*8H2O = grams/mola mass = ?
Using the coefficients in the balanced equation, convert mols Na2SO4 to mols either BaSO4 or NaOH.
Do the same and convert mols Ba(OH)2*8H2O to mols BaSO4 or NaOH (but use the same product for both calculations).It is likely that the two values will not agree; the correct value in limiting reagent problem is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.