1.) For the reaction: 2 HBr + H2SO4 --> Br2 + SO2 + 2 H2O : if you start with 5.00 g of each reactant and end up with 5.00 g of bromine, what is the percent yield?

Question

98.8%
2.0%
101.2%
100.0%

2.) For the reaction: 2 HBr + H2SO4 --> Br2 + SO2 + 2 H2O : if you start with 5.00 g of each reactant and end up with 5.00 g of bromine, what is the theoretical yield?
5.00 g
0.0510 g
4.94 g
0.06180 g

Answer 1

rounding molar masses to
HBr = 81 g/mol
Br2 = 160 g/mol
H2SO4 = 98 g/mol
so
5 g HBr = 5/81 = .0617 mol
5 g H2SO4 = 5/98 = .0510 mol
BUT I need twice as many HBr molecules as H2SO4 molecules
so the 5 g of HBr is all that matters here. I have H2SO4 left over to burn fingers or whatever.
so I should get .0617/2 = .03085 mols of Br2
.03085* 160 g/mol = 4.936 grams of Br2
That is about your 4.94 in the answer list (part 2.)
(note that only the Hydrogen mass is different between 2HBr and Br2
the percent yield = 100 * 4.936/5 = 98.7%
(unlikely :)

Answer 2

By the way you do not start up with exactly 5 g of HBr and end up with exactly 5 g of Br2. Hydrogen does not have negative mass, although it is pretty small.

Answer 3

Thank You