Question

For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of aluminum sulfate. aluminum oxide (s) + sulfuric acid (aq) aluminum sulfate (aq) + water (l) What is the theoretical yield of aluminum sulfate? grams What is the percent yield for this reaction? %

Answer 1

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For $Al_2O_3$

Mass of $Al_2O_3$  = 4.07 g

Molar mass of $Al_2O_3$  = 101.96 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4.07\ g}{101.96\ g/mol}$

$Moles\ of\ Al_2O_3= 0.0399\ mol$

According to the reaction:

$Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O$

1 mole of $Al_2O_3$  on reaction produces 1 mole of $Al_2(SO_4)_3$

So,  

0.0399 mole of $Al_2O_3$  on reaction produces 0.0399 mole of $Al_2(SO_4)_3$

Moles of $Al_2(SO_4)_3$  obtained = 0.0399 mole

Molar mass of $Al_2(SO_4)_3$ = 342.2 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0399= \frac{Mass}{342.2\ g/mol}$

$Mass= 13.7\ g$

Theoretical yield = 13.7 g

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

$\%\ yield =\frac{10.4}{13.7}\times 100$

% yield =76 %