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Chem Very Urgent

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if 4.55 g of sodium sulphide and 15.0 g of bismuth nitrate are dissolved in separate beakers of water which are then poured togehter, what is the maximum mass of bismuth sulphide an insoulble compound that could precipitate? Which reactant was the limitting reactant?

Na2S + Bi(NO3)3 --- Bi2S3 + Na^+ + NP3^- ( not balanced )



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i m getting 9.77 g for Bi2S3 .
no limiting reactant .. both are same ..
can u plz conform

  • Chem Very Urgent - ,

    No limiting reactant? Odd.

    What was the moles of Na2S?
    What was the moles of Bi(NO3)3

    Now balance the equation.
    the coefficents tells you the mole ratio you need. What was the mole ratio you had?

  • Chem Very Urgent - ,

    balanced equatoin is

    3Na2S + 2Bi(NO3)3 --- Bi2S3 + 6Na^+ + 6NO3^-

    moles of Na2S = 0.058
    moles of Bi(no3)3 = 0.038

    Na2S = 0.058 / 3 = 0.019
    Bi(no3)3 = 0.038 / 2 = 0.019

    both are same for Bi2S3 from balanced equation

    so the mass is 9.77 g

    no limiting reactant ..

    plz conform

    thanks

  • Chem Very Urgent - ,

    amazing. exactly the same. No limiting reactant.

    check your calcs again. I might have made a mistake, but I got 9.57 g.

    0.019*(2*208.98+3*32)

  • Chem Very Urgent - ,

    don't round the answers so early, and you will see that one of them is the limiting reactant

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