Question

1)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 with remain once the reaction is complete? (report in grams)

2)If the yield of the reaction in question 2 is known to be only 73.42%, how much of each of the three reagents would you need to obtain 10.000g of boron trichloride?

Answer 1

Solution :-

Balanced reaction equation

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

Molar mass    69.619           70.9        12.01         117.17          44.01

                           g/mol         g/mol        g/mol          g/mol              g/mol

If 8.254g of B₂O₃ is mixed with 4.446g of Cl₂ and 4.115g of C then

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

Solution :- lets calculate the moles of the each reactant.

Moles= mass / molar mass

Moles of B2O3 = 8.254 g / 69.619 g per mol = 0.1186 mol B2O3

Moles of Cl2 = 4.446 g / 70.9 g per mol = 0.06271 mol Cl2

Moles of C= 4.115 g C / 12.01 g per mol = 0.3673 mol C

Now using the mole ratio of the each reactant lets calculate the moles of the product BCl3

0.1186 mol B2O3 * 4 mol BCl3 / 2 mol B2O3 = 0.2372 mol BCl3

0.06271 mol Cl2 * 4 mol BCl3 /6 mol Cl2 = 0.04181 mol BCl3

0.3673 mol C * 4 mol BCl3 / 3 mol C =0.4897 mol BCl3

Moles of BCl3 produced from the Cl2 are the lowest therefore Cl2 is the limiting reactant

Therefore maximum moles of BCl3 that can be produced = 0.04181 mol BCl3

Now lets convert these moles of BCl3 to its mass

Mass = moles * molar mass

         = 0.04181 mol BCl3 * 117.17 g per mol

         = 4.90 g BCl3

Therefore the theoretical yield of the reaction = 4.90 g BCl3

B)How much of each of the excess reagents with remain once the reaction is complete? (report in grams)

Solution :- Cl2 is the limiting reactant therefore it gets used up in the reaction therefore mass of Cl2 remain = 0.0 g Cl2

Now lets calculate the mass of the C needed to react using the mole ratio with Cl2

0.04181 mol Cl2 * 3 mol C / 6 mol Cl 2 = 0.0209 mol C

So mass of C needed = 0.0209 mol * 12.01 g per mol= 0.251 g

So mass of the C remaining = 4.446 g – 0.251 g = 4.195 g C

Now lets calculate the moles of the B2O3 reacted

0.04181 mol Cl 2 * 2 mol B2O3 / 6 mol Cl2 = 0.01394 mol B2O3

So mass of B2O3 reacted = 0.01394 mol * 69.619 g per mol = 0.970 g

So mass of B2O3 remain = 8.254 g – 0.970 g = 7.284 g B2O3

2)If the yield of the reaction in question 2 is known to be only 73.42%, how much of each of the three reagents would you need to obtain 10.000g of boron trichloride?

Solution :-

Reaction gives 73.42 % yield so lets calculate the 100 % yield

10.0 g * 100 % / 73.42 % = 13.62 g BCl3

Now lets calculate the mass of each reactant needed for the reaction

13.62 g BCl3 * 1 mol / 117.17 g = 0.11624 mol

Mass of B2O3 needed is calculated as follows

(0.11624 mol BCl3 * 2 mol B2O3/4 mol BCl3)*(69.619 g/ 1 mol B2O3)= 4.05 g B2O3    

Mass of C needed is calculated as follows

(0.11624 mol BCl3 * 3 mol C / 4 mol BCl4)*(12.01 g / 1 mol C) =1.05 g C

Mass of Cl2 needed is calculated as follows

(0.11624 mol BCl3 * 6 mol Cl2/ 4 mol BCl3)*(70.9 g / 1 mol Cl2) =12.36 g Cl2