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from the reaction 4Al(s) + 3O2(g)= 2Al2O3 (s) how many grams of Al are needed to produce 50.0 g of Al2O3

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To find out how many grams of aluminum (Al) are needed to produce 50.0 grams of aluminum oxide (Al2O3), you can use stoichiometry based on the balanced chemical equation:

4Al(s) + 3O2(g) -> 2Al2O3(s)

First, you need to determine the molar mass of Al2O3 (aluminum oxide) and Al (aluminum).

Molar mass of Al2O3:

2 atoms of Al (2 × 26.98 g/mol) + 3 atoms of O (3 × 16.00 g/mol) = 53.96 g/mol + 48.00 g/mol = 101.96 g/mol

Now, you can set up a proportion using the molar mass of Al2O3 to find the moles of Al needed to produce 50.0 grams of Al2O3:

(50.0 g Al2O3) × (1 mol Al2O3 / 101.96 g Al2O3) × (4 mol Al / 2 mol Al2O3) = moles of Al

Now, calculate the moles of Al:

(50.0 / 101.96) × (4 / 2) = 2.45 moles of Al

Now that you know you need 2.45 moles of Al to produce 50.0 grams of Al2O3, you can find the mass of Al needed:

2.45 moles Al × 26.98 g/mol = 66.15 grams of Al

So, you would need approximately 66.15 grams of aluminum (Al) to produce 50.0 grams of aluminum oxide (Al2O3) in this reaction.

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