Concept:Mass-mass stoichiometry involves converting the given mass of a reactant into moles, using the balanced equation's molar ratio to find the required moles of the second reactant, and converting that back into mass.
Formula:$$ 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 $$
Solution:- First, convert 27g of Aluminum into moles. (Atomic mass of Al = 27). \( n = 27/27 = 1 \text{ mole of Al} \).
- Look at the balanced equation ratio: 4 moles of Al require 3 moles of \( \text{O}_2 \).
- Therefore, 1 mole of Al requires \( 3/4 \) moles of \( \text{O}_2 \) (which is 0.75 moles).
- Convert 0.75 moles of \( \text{O}_2 \) back into mass. (Molar mass of \( \text{O}_2 = 32 \text{ g/mol} \)).
- Mass of \( \text{O}_2 = 0.75 \text{ moles} \times 32 \text{ g/mol} = 24 \text{ grams} \).
Why other options are incorrect:- 8g: Result of assuming 1 mole of Al needs 0.25 moles of O2.
- 16g: Fails to account for the diatomic nature of Oxygen (uses 16 instead of 32 for molar mass).
- 32g: Assumes a 1:1 molar ratio, completely ignoring the 4:3 stoichiometric coefficients.
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