Question

Question 51 pts Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant, or catalyst for organic reactions. A mixture of 82.49 g of aluminum and 117.65 g of oxygen is allowed to react. What is the mass of the excess reactant present in the vessel when the reaction is complete? Report your answer to the appropriate number of significant figures.

Answer 1

Answer: 45 grams

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of
`Al`

`\text{Number of moles}=(82.49g)/(27g/mol)=3.1moles`

b) moles of
`O_2`

`\text{Number of moles}=(117.65g)/(32g/mol)=3.7moles`

`4Al+3O_2\rightarrow 2Al_2O_3`

According to stoichiometry :

4 moles of
`Al` require = 3 moles of
`O_2`

Thus 3.1 moles of
`Al` will require=
`(3)/(4)* 3.1=2.3moles` of
`O_2`

Thus
`Al` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent as (3.7-2.3)= 1.4 moles of
`O_2` will remain unreacted.

Mass of
`O_2=moles* {\text {Molar mass}}=1.4moles* 32g/mol=45g`

Thus 45 g of
`O_2` will be present in the vessel when the reaction is complete.