posted by Matt

When 0.500 moles of boron trichloride react with 1.20 moles hydrogen gas to produce elemental boron and hydrogen chloride gas, the actual yield of boron was 66.4 % of the theoretical yield. The mass of boron obtained was _____ g.

Please help I'm stuck on it.

  1. DrBob222

    2BCl3 + 3H2 ==> 2B + 6HCl
    0.5 mol BCl2 will produce 0.5 mol B if you have all of the H2 needed.
    1.20 mols H2 gas will produce 1.20 x (2 mol B/3 mols H2) = 1.20 x 2/3 = 0.8 mol B if you hadd all of the BCl3 needed.
    Both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. In this case the LR is BCl3 and H2 is the excess reagent so you can produce 0.5 mol B as the theoretical yield. 0.5 mols x atomic mass B = estimated 5 g B. The yield is only 66.4%; therefore, you can collect only estimated 5 x 0.664 = ? g B.

  2. Matt

    Can we use moles of HCl instead of B to calculate limiting reactant?

  3. DrBob222

    Yes. The limiting reagent will always produce the LEAST product (whichever one you choose)

  4. Anonymous


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