Respuesta :
Answer:
18.82L of oxygen gas are needed.
Explanation:
1st) It is necessary to write and balance the chemical reaction:
[tex]3O_2+4Al\rightarrow2Al_2O_3[/tex]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:
[tex]\begin{gathered} 1mol-32g \\ 3mol-x=\frac{3mol*32g}{1mol} \\ x=96g \end{gathered}[/tex]- Al conversion:
[tex]\begin{gathered} 1mol-27g \\ 4mol-x=\frac{4mol*27g}{1mol} \\ x=108g \end{gathered}[/tex]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:
[tex]\begin{gathered} 108gAl-96gO_2 \\ 30.25gAl-x=\frac{30.25gAl*96gO_2}{108gAl} \\ x=26.89gO_2 \end{gathered}[/tex]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:
[tex]\begin{gathered} 1mol-22.4L \\ 0.84mol-x=\frac{0.84mol*22.4L}{1mol} \\ x=18.82L \end{gathered}[/tex]So, 18.82L of oxygen gas are needed.
