Question

How many liters of oxygen gas at STP are required to react with 30.25 g of aluminum in the production of aluminum oxide?

Answer 1

18.82L of oxygen gas are needed.

Step-by-step explanation:

1st) It is necessary to write and balance the chemical reaction:

`3O_2+4Al\rightarrow2Al_2O_3`

2nd) From the balanced reaction, we can see that 3 moles of oxygen gas (O2) react with 4 moles of aluminum (Al). To convert moles to grams, it is necessary to use the molar mass of oxygen (32g/mol) and aluminum (27g/mol):

- O2 conversion:

`\begin{gathered} 1mol-32g \\ 3mol-x=(3mol*32g)/(1mol) \\ x=96g \end{gathered}`

- Al conversion:

`\begin{gathered} 1mol-27g \\ 4mol-x=(4mol*27g)/(1mol) \\ x=108g \end{gathered}`

Now we can see that 96g of O2 react with 108g of Al.

3rd) We have to calculate the grams of O2 that will react with 30.25g of Al:

`\begin{gathered} 108gAl-96gO_2 \\ 30.25gAl-x=(30.25gAl*96gO_2)/(108gAl) \\ x=26.89gO_2 \end{gathered}`

Using the molar mass of oxygen, we know that 26.89g represent 0.84 moles of O2.

4th) Finally, a mole of a gas at STP conditions occupies a volume of 22.4L. With this number and the moles of oxygen gas, we can calculate the liters:

`\begin{gathered} 1mol-22.4L \\ 0.84mol-x=(0.84mol*22.4L)/(1mol) \\ x=18.82L \end{gathered}`

So, 18.82L of oxygen gas are needed.