Chemistry: Balance equation, find volume of produce, and percent yield.

Question

B2O3(s)+ C(s)+Cl2(g)=BCl3(l)+CO(g)

1.) Write the balanced molecular reaction. Include phases.

2.) Assume that excess carbon is present .0.456 g of B2O3 reacts with 7.89 L of Cl2(g) at standard temperature and pressure. The products are collected at 23.2 C with a pressure of 1.03 bar. what volume of CO(g) is produced?

3.)If 1.00 L of CO(g) are collected, what is the percent yield for this reaction? State one reason why your percent yield is greater than or less than 100%.

Answer 1

1). B₂O₃(s) + 3C(s) + 3Cl₂(g )==> 2BCl₃(l) + 3CO(g) ... balanced equation

2). Find the limiting reactant by finding moles of B2O3 and moles of Cl2:

moles B2O3 = 0.456 g B2O3 x 1 mol/69.6 g = 0.00655 moles B2O3

moles Cl2: use PV = nRT ... n = PV/RT where P = 1.03 bar x 1 atm/1.01325 bar = 1.02 atm, V = 7.89 L, R = 0.0821 Latm/Kmol and T = 23.2 + 273 = 296K. n = (1.02)(7.89)/(0.0821)(296) = 0.331 moles Cl2

So, clearly the B2O3 is limiting even though you need 3 mol Cl2 per mol B2O3 there is excess B2O3.

Now, we can find the volume of CO produced by finding moles of CO2 and then converting to volume.

moles CO = 0.00655 mol B2O3 x 3 mol CO/mol B2O3 = 0.0197 mol CO produced

From PV = nRT we have V = nRT/P

Volume CO = (0.0197)(0.0821)(296)/(1.02) = 0.469 L of CO produced

3). The % yield would be actual/theoretical (x100%) = 1.00 L/0.469 L (x100%) = 213%

One reason for this being greater than 100% may be that the Cl2 gas is also contributing to the total pressure so you'd want to use the partial pressure of CO to find the percent yield.