Question

Boron trichloride is prepared from the following reaction. 2B2O3 + 6Cl2 + 3C --> 4BCl3 + 3CO2

If 8.254g of B2O3 is mixed with 4.446g of Cl2 and 4.115g of C then answer the following questions.

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

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

Answer 1

2B2O3 + 6Cl2 + 3C --> 4BCl3 + 3CO2

2 x 69.62 g B2O3 reacts with 6 x 71 g Cl2

8.254 g B2O3 reacts with 8.254 x 6 x 71 / 2 x 69.62 = 3516.204 / 139.24 = 25.2 g Cl2

but we have only 4.446 g Cl2

so Cl2 li limiting reagent

6 x 71 g Cl2 forms 4 x 117.17 g BCl3

4.446 g Cl2 forms 4.446 x 4 x 117.17 / 6 x 71 = 2083.75 / 426 = 4.89 g

therotical yield of BCl3 = 4.89 g

b) 6 x 71 g Cl2 reacts with 3 x 12 g C

4.446 g Cl2 reacts with 4.446 x 3 x 12 / 6 x 71 = 0.376 g

exess C left = 4.115 - 0.376 = 3.739 g

6 x 71 g Cl2 reacts with 2 x 69.62 g B2O3

4.446 g Cl2 reacts with 4.446 x 2 x 69.62 / 6 x 71 = 1.45 g B2O3

exess B2O3 left = 8.254 - 1.45 = 6.804 g