Concept:Mass-mass stoichiometry allows us to predict the mass of a product formed by converting the given reactant mass to moles, applying the molar ratio from the balanced equation, and converting back to mass.
Formula:$$ 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 $$
Solution:- First, calculate the moles of Aluminium (Al) supplied:
- Atomic mass of Al = \( 27 \text{ g/mol} \).
- Moles of Al = \( \frac{18.5 \text{ g}}{27 \text{ g/mol}} \approx 0.685 \text{ moles} \).
- According to the balanced equation, 4 moles of Al produce exactly 2 moles of \( \text{Al}_2\text{O}_3 \). This is a 4:2 (or 2:1) ratio.
- Therefore, the moles of \( \text{Al}_2\text{O}_3 \) produced will be exactly half the moles of Al consumed: \( 0.685 / 2 = 0.3425 \text{ moles} \).
- Now, convert moles of \( \text{Al}_2\text{O}_3 \) back into mass. Molar mass of \( \text{Al}_2\text{O}_3 = (2 \times 27) + (3 \times 16) = 54 + 48 = 102 \text{ g/mol} \).
- Mass = \( 0.3425 \text{ moles} \times 102 \text{ g/mol} = 34.935 \text{ grams} \).
- Rounding to one decimal place yields \( 34.9 \text{ g} \).
Why other options are incorrect:- 30.8g, 32.6g, 36.5g: These incorrect values arise from bypassing the 4:2 molar ratio (e.g., assuming a 1:1 conversion) or using an incorrect molar mass for Aluminium Oxide.
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