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Answer :
Final answer:
The question asks how many grams of Al2O3 can be produced from 34.7 grams of O2. After converting grams of O2 to moles, using the molar ratio from the reaction 4 Al + 3 O2 ightarrow 2 Al2O3, and converting moles of Al2O3 to grams, the theoretical yield of Al2O3 is found to be 73.7 grams, which does not match the given options. So , the correct option is a) 73.7 grams
Explanation:
To determine how many grams of Al2O3 can be produced from 7 grams of O2 , we first need to look at the balanced chemical equation for the reaction between aluminum and oxygen to form aluminum Unfortunately, that equation is not provided, but the typical reaction is:
4 Al + 3 O2 ightarrow 2 Al2O3
We can then use stoichiometry to find the amount of Al2O3 produced from O2. First, we convert grams of O2 to moles, and then use the molar ratio from the equation to find moles of Al2O3. However, this requires the molar mass of O2 and Al2O3, which we will assume to be 00 g/mol for O2 and 96 g/mol for Al2O3 based on periodic table values:
- Determine moles of O2: (7 / (00 g/ = 0844 mol O2
- Use molar ratio (3 moles O2 for 2 moles Al2O3): (2 moles Al2O3 / 3 moles O2) imes 0844 mol O2 = 7229 mol Al2O3
- Convert moles of Al2O3 to grams: 7229 mol imes 96 g/mol = 7 grams of Al2O3 (rounded to one decimal
This calculated value does not match any of the provided 6 grams, 4 grams, 5 grams, or 3 grams, indicating a possible discrepancy in the question's values or the provided Chemistry calculations must be based on accurate molar masses and stoichiometric If further context was provided, such as the correct balanced equation and molar masses, a more accurate answer could be
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