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

Theoretical yield = 13.7 g

% yield =76 %

Step-by-step 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`0.0399= (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}* 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 =(10.4)/(13.7)* 100`

% yield =76 %