Answer to Question #205172 in General Chemistry for emma

Question

2AlBr3 + 3K2SO4 = 6KBr + Al2(SO4)3

a) A student combines 12.7 grams of AlBr3 with 18.2 grams of K2SO4. How many grams of KBr will be produced from this reaction?

b)  In a separate experiment using these same reactants, a student determined that theoretically, they should have produced 157 grams of Al2(SO4)3 however when the experiment was complete they only produced 145 grams. What is the percent yield? 

Answer 1

Balanced equation:

2 AlBr3 + 3 K2SO4 \rightarrow 6 KBr + Al2(SO4)3

Mass of AlBr3 = 7.145 g

Molecular weight of AlBr3 = 266.69 gmol-1

Moles of AlBr3 = 7.145 g / 266.69 gmol-1 = 0.02679 Moles

Mass of K2SO4 = 6.839 g

Molecular weight of K2SO4 = 174.25 gmol-1

Moles of K2SO4 = 6.839 g / 266.69 gmol-1 = 0.03924 Moles

K2SO4 should be a limiting reagent

Theoretical moles of KBr = 0.03924 x 6/3 = 0.07849 moles

Theoretical mass of KBr = 0.07849 mol x 119 gmol-1 = 9.340 g

Actual yield of KBr obtained = 8.721 g

% yield = \frac{actualyield }{theoretical yield} x 100 %

Percentage yield = \frac{8.721g x100 }{9.34g} = 93.37​​​​​​​%

Hence the % yield of the reaction is 93.37 %