Question

How many grams of oxygen will be required to react completely with 27g of Al

Answer 1

Answer:

prerequisite knowledge is that number of moles of any substance is equal to to the ratio of of given mass and gram molecular mass

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Answer 2

Answer: The mass of oxygen gas required to react is 48 grams

Explanation:

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

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

Given mass of aluminium = 27 g

Molar mass of aluminum = 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 aluminium and oxygen follows:

$2Al+3O_2\rightarrow 2Al_2O_3$

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of oxygen gas

So, 1 mole of aluminium will react with = $\frac{3}{2}\times 1=1.5mol$ of oxygen gas

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

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 1.5 moles

Putting values in equation 1, we get:

$1.5mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(1.5mol\times 32g/mol)=48g$

Hence, the mass of oxygen gas required to react is 48 grams