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?
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
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 :)
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.
Thank You
To calculate the percent yield and theoretical yield for a chemical reaction, we need to understand the concept of stoichiometry and the mole-to-mole ratio between reactants and products.
1.) To find the percent yield, we compare the actual (experimental) yield to the theoretical yield and express it as a percentage:
Theoretical Yield = Stoichiometric ratio * Limiting reactant mass
In this case, the limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. We need to determine which reactant is limiting by calculating the amount of bromine that can be produced from each reactant. Since we start with 5.00 g of each reactant, we can calculate the amount of bromine that can be produced from each:
From 2 HBr: (5.00 g HBr) * (1 mol HBr / molar mass HBr) * (1 mol Br2 / 2 mol HBr) * (molar mass Br2) = X g Br2
From H2SO4: (5.00 g H2SO4) * (1 mol H2SO4 / molar mass H2SO4) * (1 mol Br2 / 1 mol H2SO4) * (molar mass Br2) = Y g Br2
The actual yield is given as 5.00 g of bromine. To calculate the percent yield, divide the actual yield by the theoretical yield and multiply by 100:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Calculate the theoretical yields from both reactants and compare them to 5.00 g of bromine to determine the percent yield.
2.) The theoretical yield is the maximum amount of product that can be formed based on the stoichiometry of the reaction. It is calculated using the same approach as in question 1, but without comparing it to the actual yield.
Calculate the theoretical yield of bromine by determining the limiting reactant and using its stoichiometric ratio to convert the mass of the limiting reactant to the mass of bromine produced.