Question

This question involves two calculations. The answer to the first part will be

Answer 1

Answer :

(a) The mass of $Al_2O_3$ produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of $Al$ = 85.1 g

Molar mass of $Al$ = 27 g/mol

First we have to calculate the moles of $Al$

$\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol$

Now we have to calculate the moles of $Al_2O_3$

The balanced chemical equation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the reaction, we conclude that

As, 4 moles of $Al$ react to give 2 moles of $Al_2O_3$

So, 3.15 moles of $Al$ react to give $\frac{2}{4}\times 3.15=1.58$ mole of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

Molar mass of $Al_2O_3$ = 102 g/mole

$\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g$

Therefore, the mass of $Al_2O_3$ produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%$

Therefore, the percent yield of the reaction is, 72.5 %