Calculating the Weight of Oxygen Required to React with Aluminum
Understanding the Reaction:
- The reaction between aluminum (Al) and oxygen (O₂) forms aluminum oxide (Al₂O₃).
- The balanced chemical equation for this reaction is 4Al + 3O₂ → 2Al₂O₃.
Given Information:
- Mass of aluminum (Al) = 27 grams.
Calculating the Molar Mass of Aluminum (Al):
- The molar mass of aluminum (Al) is 27 grams/mol (based on its atomic mass from the periodic table).
Determining the Mole Ratio:
- From the balanced chemical equation, we see that 4 moles of aluminum react with 3 moles of oxygen.
- This gives us a mole ratio of 4:3 between aluminum and oxygen.
Calculating the Moles of Aluminum:
- To find the moles of aluminum, we divide the given mass by the molar mass: 27g / 27g/mol = 1 mole.
Calculating the Moles of Oxygen:
- Using the mole ratio, we determine the moles of oxygen required: 1 mole of Al requires 3/4 moles of O₂.
- Therefore, 1 mole of Al requires 3/4 moles of O₂.
Calculating the Weight of Oxygen:
- To find the weight of oxygen required, we multiply the moles of O₂ by its molar mass: (3/4) * 32g/mol = 24g.
- Therefore, 24 grams of oxygen is required to completely react with 27 grams of aluminum.
In conclusion, 24 grams of oxygen are needed to completely react with 27 grams of aluminum to form aluminum oxide.