Given the reaction 4Al + 3O_2 → 2Al_2O_3, how many grams of aluminum (Al) are required to produce 3.5 moles of Al_2O_3 in the presence of excess O_2?

• A. 94.5 g
• B. 158.2 g
• C. 107.8 g
• D. 189.0 g

Answer: (B)

Verified step by step guidance

1. Identify the balanced chemical equation: \$4Al + 3O_2 \rightarrow 2Al_2O_3$. This tells us the mole ratio between aluminum and aluminum oxide.
2. Determine the mole ratio between aluminum and aluminum oxide from the balanced equation. For every 2 moles of \(Al_2O_3\) produced, 4 moles of \(Al\) are required. So, the mole ratio is \(\frac{4 \text{ moles } Al}{2 \text{ moles } Al_2O_3} = 2\) moles of \(Al\) per mole of \(Al_2O_3\).
3. Calculate the moles of aluminum needed to produce 3.5 moles of \(Al_2O_3\) using the mole ratio: \$3.5 \text{ moles } Al_2O_3 \times 2 \frac{\text{moles } Al}{\text{mole } Al_2O_3} = 7\( moles of \)Al$.
4. Find the molar mass of aluminum (Al) from the periodic table, which is approximately 26.98 g/mol.
5. Calculate the mass of aluminum required by multiplying the moles of aluminum by its molar mass: \$7 \text{ moles } Al \times 26.98 \frac{g}{mol} = $ mass of aluminum in grams.