Answer:
18.82L of oxygen gas are needed.
Explanation:
1st) It is necessary to write and balance the chemical reaction:
![3O_2+4Al\rightarrow2Al_2O_3](https://tex.z-dn.net/?f=3O_2%2B4Al%5Crightarrow2Al_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=\frac{3mol*32g}{1mol} \\ x=96g \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%201mol-32g%20%5C%5C%203mol-x%3D%5Cfrac%7B3mol%2A32g%7D%7B1mol%7D%20%5C%5C%20x%3D96g%20%5Cend%7Bgathered%7D)
- Al conversion:
![\begin{gathered} 1mol-27g \\ 4mol-x=\frac{4mol*27g}{1mol} \\ x=108g \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%201mol-27g%20%5C%5C%204mol-x%3D%5Cfrac%7B4mol%2A27g%7D%7B1mol%7D%20%5C%5C%20x%3D108g%20%5Cend%7Bgathered%7D)
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=\frac{30.25gAl*96gO_2}{108gAl} \\ x=26.89gO_2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20108gAl-96gO_2%20%5C%5C%2030.25gAl-x%3D%5Cfrac%7B30.25gAl%2A96gO_2%7D%7B108gAl%7D%20%5C%5C%20x%3D26.89gO_2%20%5Cend%7Bgathered%7D)
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=\frac{0.84mol*22.4L}{1mol} \\ x=18.82L \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%201mol-22.4L%20%5C%5C%200.84mol-x%3D%5Cfrac%7B0.84mol%2A22.4L%7D%7B1mol%7D%20%5C%5C%20x%3D18.82L%20%5Cend%7Bgathered%7D)
So, 18.82L of oxygen gas are needed.
Nuclear energy
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Answer:
identify and name the landforms in the following diagrams
Answer:
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