Aluminum and oxygen are mixed together to yield aluminum oxide. If 3.17 g of Al are used, how many g of product can be made? Al+ O2 yields Al2O3
the balanced equation is
4Al + 3O2 = 2Al2O3
so, how many moles of Al do you have?
You get half that many moles of Al2O3
now convert back to grams
What is the answer, I want to know if I am right.
nice try. tell me your answer, and I'll check it out.
To determine the amount of product that can be made, we need to use the concept of stoichiometry, which is the study of the quantitative relationship between reactants and products in a chemical reaction.
1. Start by writing down the balanced chemical equation for the reaction:
4 Al + 3 O2 → 2 Al2O3
2. Determine the molar mass of aluminum (Al) and aluminum oxide (Al2O3):
- Aluminum (Al): 26.98 g/mol
- Aluminum oxide (Al2O3): 101.96 g/mol
3. Convert the given mass of aluminum (3.17 g) to moles by dividing by its molar mass:
Moles of Al = Mass of Al / Molar mass of Al
= 3.17 g / 26.98 g/mol
≈ 0.1175 mol
4. Use the balanced chemical equation to determine the mole ratio between aluminum (Al) and aluminum oxide (Al2O3):
From the equation: 4 Al : 2 Al2O3
Therefore, 1 Al : 1/2 Al2O3
5. Calculate the moles of aluminum oxide (Al2O3) that can be formed by multiplying the moles of aluminum (0.1175 mol) by the mole ratio:
Moles of Al2O3 = Moles of Al × (1/2)
= 0.1175 mol × (1/2)
= 0.05875 mol
6. Finally, convert the moles of aluminum oxide (0.05875 mol) to grams by multiplying by its molar mass:
Mass of Al2O3 = Moles of Al2O3 × Molar mass of Al2O3
= 0.05875 mol × 101.96 g/mol
≈ 5.99 g
Therefore, approximately 5.99 grams of aluminum oxide (Al2O3) can be produced when 3.17 grams of aluminum (Al) react with oxygen (O2).