Question

2)Boron trichloride is prepared from the following reaction. 2B2O3 + 6Cl2 + 3C --> 4BCl3 + 3CO2 If 3.456g of B2O3 is mixed with 9.9216g of Cl2 and 2.459g of C then answer the following questions.

2a) What is the theoretical yield of boron trichloride in grams?

2b) How much of each of the excess reagents would remain once the reaction is complete? (report in grams)

3) If the yield of the reaction in question 2 is known to be only 67.89% using stochometric amounts, how much (g) of each of the three reagents would you need to obtain 10.000g of boron trichloride?

Answer 1

For the given reaction,

a) moles B2O3 = 3.456 g/69.62 g/mol = 0.05 mol

moles Cl2 = 9.9216 g/71 g/mol = 0.14 mol

moles C = 2.459 g/12 g/mol = 0.205 mol

moles Product formed starting with B2O3 = 0.05 x 4/2 = 0.10 mol

moles Product formed starting with Cl2 = 0.14 x 4/6 = 0.093 mol

moles Product formed starting with C = 0.205 x 4/3 = 0.273 mol

So limiting reactant = Cl2

theoretical yield of BCl3 = 0.093 mol x 117.17 g/mol = 10.9 g

b) excess reagent B2O3 left after reaction = 0.05-(0.14 mol x 2/6) x 69.62 g/mol = 0.232 g

excess reagent C left after reaction = 0.205-(0.14 mol x 3/6) x 12 g/mol = 1.62 g

c) yield is 67.89%

BCl3 yield = (10 g/117.17 g/mol) = 0.0853 mol

0.0853 mol is 67.89% so, 100% = 0.0853 x 100/67.89 = 0.126 mol

mass B2O3 needed = 0.126 mol x 2 x 69.62 g/mol/4 = 4.386 g

mass Cl2 needed = 0.126 mol x 6 x 71 g/mol/4 = 13.419 g

mass C needed = 0.126 mol x 3 x 12 g/mol/4 = 1.134 g