Answer to Question #205172 in General Chemistry for emma

Question #205172

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? 


1
Expert's answer
2021-06-10T03:32:18-0400

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 %


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS