Chemistry

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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?

  • Chemistry -

    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.

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