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.

Question

Please help I'm stuck on it.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

3.59