When 27 grams of Al reacts completely with excess O_2, how many moles of Al_2O_3 are formed?

• A. 1.5 moles
• B. 0.5 moles
• C. 1.0 mole
• D. 2.0 moles

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between aluminum (Al) and oxygen (O_2) to form aluminum oxide (Al_2O_3): \[2Al + \frac{3}{2}O_2 \rightarrow Al_2O_3\]
2. Calculate the number of moles of aluminum (Al) given its mass. Use the molar mass of Al, which is approximately 27 g/mol: \[\text{moles of Al} = \frac{27\ \text{g}}{27\ \text{g/mol}}\]
3. Use the stoichiometric ratio from the balanced equation to find the moles of aluminum oxide (Al_2O_3) produced. According to the equation, 2 moles of Al produce 1 mole of Al_2O_3, so: \[\text{moles of } Al_2O_3 = \frac{1}{2} \times \text{moles of Al}\]
4. Substitute the moles of Al calculated in step 2 into the stoichiometric ratio from step 3 to find the moles of Al_2O_3 formed.
5. Interpret the result to determine which answer choice corresponds to the calculated moles of Al_2O_3.