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How many grams of Al₂O₃ can form from 38.2 g of Al?

Reaction:
\[ 4 \text{Al}(s) + 3\text{O}_2(g) \rightarrow 2 \text{Al}_2\text{O}_3(s) \]

Step: Show the strategy for solving this problem.

Answer :

Final answer:

To calculate the mass of Al2O3 from 38.2 g of Al, convert the mass of aluminum to moles, use the stoichiometry of the reaction to find moles of Al2O3 produced, and then convert those moles back to grams using the molar mass of Al2O3.

Explanation:

To calculate the mass of aluminum oxide (Al2O3) that can be formed from 38.2 g of aluminum (Al), we must first use the molar mass of aluminum to convert mass to moles. Then, we'll use the stoichiometry of the balanced reaction to find the number of moles of Al2O3 that can be produced. After that, we can convert moles of Al2O3 back to grams by using its molar mass.

The molar mass of Al is approximately 26.98 g/mol. The balanced equation is 4 Al(s) + 3 O2(g) → 2 Al2O3(s). According to the equation, 4 moles of Al produce 2 moles of Al2O3. So, for every mole of Al, half a mole of Al2O3 is produced.

Step 1: Calculate the moles of Al:

Moles of Al = Mass of Al ÷ Molar mass of Al

Moles of Al = 38.2 g ÷ 26.98 g/mol

Step 2: Using the stoichiometry of the reaction, calculate moles of Al2O3:

Moles of Al2O3 = Moles of Al × (1/2)

Step 3: Convert moles of Al2O3 to mass:

The molar mass of Al2O3 is approximately 101.96 g/mol.

Mass of Al2O3 = Moles of Al2O3 × Molar mass of Al2O3

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