Question

1) If you had not dried the solid, would the calculated percent yield have been higher, lower, or unchanged? Explain

2)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.   

- What is the theoretical yield of boron trichloride in grams?

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

3) 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

(1) The wet solid contains the solvent molecules which are takes the interstitial positions of the solid compound there by the mass of the wet sample is equal to mass of solid plus mass of the solvent. So the mass of the wet sample is more ,therefore the percentage yield is higher than the actual value.

(2) 2B₂O₃ + 6Cl₂ + 3C 4BCl₃ + 3CO₂

Molar mass of B₂O₃ = (2x11) + (3x16) = 70 g/mol

Molar mass of Cl₂ = 2x35.5 = 71 g/mol

Molar mass of C = 12 g/mol

According to the balanced equation ,

2 moles of B₂O₃ reacts with 6 moles of Cl₂ & 3 moles of C

                                         OR

2 x70 =140 g of B₂O₃ reacts with 6x71 = 426 g of Cl₂ & 3x12=36 g of C

M g of B₂O₃ reacts with 4.446 g of Cl₂ & N g of C

M = (140x4.446)/426

   = 1.46 g of B₂O₃

Mass of B₂O₃ left unreacted = 8.254 - 1.46 = 6.794 g

N = (36x4.446)/426

    = 0.36 g of C

Mass of C left unreacted is = 4.115 - 0.36 = 3.755 g

Since all the mass of Cl₂ reacts completly ,Cl₂ is the limiting reactant.

Molar mass of BCl₃ = 11 + (3x35.5) = 117.5 g/mol

From the balanced equation ,

6 moles of Cl₂ produces 4 moles of BCl₃

                         OR

6x71= 426 g of Cl₂ produces 4x117.5 = 470 g of BCl₃

4.446g of Cl₂ produces Y g of BCl₃

Y = (470 x 4.446 ) / 426

   = 4.90 g of BCl₃

Therefore the theoretical yield of BCl₃ is 4.90 g