Question

The weight of oxygen required to completely react with 27 grams of aluminium is

Answer 1

this is answer for your question

Answer 2

Answer: The mass of oxygen gas required is 24 grams.

Explanation:

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

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

Given mass of aluminium = 27 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{27g}{27g/mol}=1mol$

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

$4Al+3O_2\rightarrow 2Al_2O_3$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas.

So, 1 mole of aluminium will react with = $\frac{3}{4}\times 1=0.75mol$ of oxygen gas.

Now, calculating the mass of oxygen by using equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.75 mol

Putting values in equation 1, we get:

$0.75mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=24g$

Hence, the mass of oxygen gas required is 24 grams.