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 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

1. since the solid is not dried, the mass reads more and hence percent yield will be higher due to excess water present in the undried sample.

2. The reaction is 2B2O3+6Cl2+3C----4BCl3+ 6CO2, the molar ratio of Reactants B2O3: Cl2 : C= 2: 6:3= 1: 3:1.5 ( theoretical ratio)

moles = mass/ molar mass , molar masses : B2O3= 69, Cl2= 71 and C= 12

moles : B2O3= 3.456/69= 0.05, Cl2= 9.9216/71= 0.14, C= 2.459/12= 0.20actual molar ratio of B2O3: Cl2: C= 0.05: 0.14 :0.20, dividing by the lower value of 0.05

the ratio of B2O3: Cl2: C= 0.05/0.05 :0.14/0.05 :0.20/0.05= 1:2.8 : 4

the limiting reactant is Cl2, since required is 3 where as avaialble is only 2.8. 4 C is aslo excess and 1 of B2O3 is also excess.

moles of Cl2= 0.14 moles, moles of B2O3 required = (1/3)*0.14= 0.047, moles available=0.05, moles of C to be used as per the reaction= (3/6)*0.14= 0.07 , moles of C available= 0.2, moles of B2O3 remaining =0.05-0.047= 0.003, mass of B2O3 remaining = 0.003*69= 0.207 gm, C remaining = 0.2-0.07= 0.13, mass of C remaning = 0.13*12= 1.56 gm

moles of BCl3 formed= (4/6)*0.14= 0.093, mass of BCl3 formed= moles* molar mass of BCl3= 0.03*117.2= 10.9 g

Yield is 67.89%, 100* actual yield/ theoretical yield= 67.89

actual yield= 67.89*10.9/100 =7.4 gm

moles of BCl3 formed = 7.4/117.2= 0.063,

hence moles of B2O3 consumed = 0.063/2 =0.0315, mass of B2O3= 0.0315*69=2.2 gm

moles of Cl2 consumed = (6/4)*0.0315= 0.04725, mass of Cl2 used= 0.04725*71=3.35 gm

mass of C consumed= (3/4)*0.0315= 0.024, mass of C used= 0.024*12= 0.288 gm

7.4 gm of BCl3 requires 2.2 gm of B2O3, 3.35 gm of Cl2 and 0.288 gm of C

10 gm of BCl3 required 10*2.2/7.4= 3 gm of B2O3. 3.35*10/7.4 = 4.53 gm of Cl2 and 10.74*0.288/7.4=0.42 gm of C