Question

Calculate the number of grams of Al2O3 that could be produced if 2.5 g of aluminum and 2.5 g of oxygen were allowed to react according to the following balanced equation.

Answer 1

Answer: The mass of $Al_2O_3$ is 4.7 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

For aluminium:

Given mass of aluminium= 2.5 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

Given mass of oxygen gass = 2.5 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas:}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of aluminium and oxygen gas follows:

$4Al+3O_2\rightarrow 2Al_2O_3$

By Stoichiometry of the reaction:

4 moles of aluminium react with = 3 moles of oxygen

So, 0.092 moles of aluminium react with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produces 2 moles of aluminium oxide

So, 0.092 moles aluminium produce = $\frac{2}{4}\times 0.092=0.046moles$ of aluminium oxide

Now, calculating the mass of aluminium oxide${\text{Mass of of aluminium oxide}}=moles\times {\text{Molar mass of aluminium oxide}}=0.046mol\times 102g/mol=4.7g$

Hence, the mass of $Al_2O_3$ is 4.7 grams