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How much solid product formed with 30ml of NaOH and 40mL of NiCl2?

I'm not really sure how to go about this?

  • incomplete---Chemistry -

    Can't be done. Need concentrations of NaOH and NiCl2.

  • Chemistry -

    They are both 1Mole

  • Chemistry -

    mole is not a concentration. You may mean 1 M which is one (1) molar.
    2NaOH + NiCl2 ==> 2NaCl + Ni(OH)2

    mols NaOH = M x L = 1M x 0.030 = 0.030
    mols NiCl2 = 1M x 0.040 = 0.040

    Now use the coefficients in the balanced equation to convert mols NaOH to mols Ni(OH)2 and mols NiCl2 to mols Ni(OH)2.

    First, convert NaOH.
    mols Ni(OH)2 = 0.03 mols NaOH x (1 mol Ni(OH)2/2 mols NaOH) = 0.03 x (1/2) = 0.015 mols Ni(OH)2.

    Next, convert NiCl2.
    mols Ni(OH)2 = 0.040 x (1 mol Ni(OH)2/1 mol NiCl2) = 0.04 x 1/1 = 0.04.

    You have obtained two answers for mols Ni(OH)2 produced. Both can't be right; in limiting reagent problems (and this is one) the smaller answers is always the correct one and the reagent producing that value is the limiting reagent. Therefore, there will be 0.015 mol Ni(OH)2 produced. If you want grams covert that value. g = mols x molar mass.

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