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.   

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

2b) 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) If you haven't dried the sample, there is water/moisture in the sample, therefore the sample will have greater mass. The actual percentage yield will be very optimistic,that is, very high, compared vs. the real theoretical value

2)

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.   

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

Change everything to moels, we need MW

MW of B2O3

MW of CO2 = 44

MW of Cl2 = 70

MW of B2O3 =69.62

MW of C =12

Change to moles (moles = mass/MW)

8.254g of B2O3 --> mol = 8.254/69.62= 0.1185

4.446g of Cl2 --> mol = 4.446/70 = 0.0635

4.115g of C -> mol = 4.115/12 = 0.3429

Recall that equation requires

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

0.1185 0.0635   0.3429

From here, find the limiting reactant... clearly is Cl2 gas since it is too litle and requires

0.0635 mol of Cl will react to form:

1/2 of C

2/6 of B2O3

4/6 of BCl3

Therefore

4/6*0.0635 = 0.0423 mol of BCl3

Theoretical yield is mass, not moles so:

mass = mol*MW = 0.0423*117.17 = 4.956 g of BCl3

The theoretical yield is 4.956 g of BCl3

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

Excess reagent:

first calculate the amount of moles that did not reacted

0.0635 mol of Cl will react to form:

1/2 of C --> 0.0635 /2 = 0.03175

2/6 of B2O3 --> 1/3*0.0635 = 0.02116

That is what reacted! we need what is left!

B2O3 = 0.1185 - 0.02116 = 0.09734

C = 0.3429 - 0.03175 = 0.311

we need this in grams so:

0.09734 mol of B2O3 --> mol*MW = 0.09734 *69.6182 = 6.7766 g of B2O3

0.311 mol of C --> mol*MW = 0.311*12 = 11.689 g of C

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?

if yield is only 73.42% from that original...

how much do you need of the 3 reagents to react 10 grams of BCl3

2B2O3 + 6Cl2 + 3C --> 4BCl3 + 3CO2   

m = 10 g of BCl3

MW of BCl = 117.17 g

moles of BCl required = 10 / 117.17 = 0.5824 mol of BCl3 required

But we have a yield of 73.42 that is 0.7342

Therefore

0.5824 / 0.7342 = 0.7932 will be the "target"

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

Ratios are given as follow:

0.7932 *2 mol of B2O3 = 1.586 mol

0.7932 * 1.5 mol of Cl2 = 1.1898 mol of Cl2

0.7932 *1.333 mol of C = 1.057 mol of C