27g of Aluminum reacts with how much O2 to produce Aluminum oxide?
Correct answer: C. 0.75 moles
- A. 3 moles
- B. 0.27 moles
- C. 0.75 moles
- D. 1.5 mole
Explanation
First, write a balanced chemical equation for the reaction 4Al+ 3O2 ---> 2Al2O3 Calculate the moles in 27g Al (we know that Mr of Al is 27) mol = mass/Mr = 27/27 = 1 mol The molar ratio of Al and O2 is 4 : 3 So, the moles of O2 = (3 x 1)/ 4 = 0.75 mol
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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