4Al + 3O2 —> 2Al2O3 The molar mass of O2 is 32.0 g/mol. What mass, in grams, of O2 must react to form 3.80 mol of Al2O3? A.) 60.8 B.) 81.1 C.) 122 D.) 182
The Correct Answer and Explanation is:
Correct Answer: D) 182 grams
Step-by-Step Solution:
The balanced chemical equation is:
4Al + 3O₂ → 2Al₂O₃
Step 1: Find the mole ratio between O₂ and Al₂O₃
From the equation:
- 3 moles of O₂ produce 2 moles of Al₂O₃
So, to find how many moles of O₂ are needed to produce 3.80 moles of Al₂O₃, set up the ratio:3 moles O₂2 moles Al₂O₃=x moles O₂3.80 moles Al₂O₃\frac{3 \text{ moles O₂}}{2 \text{ moles Al₂O₃}} = \frac{x \text{ moles O₂}}{3.80 \text{ moles Al₂O₃}}2 moles Al₂O₃3 moles O₂=3.80 moles Al₂O₃x moles O₂
Step 2: Solve for xx=3×3.802=5.70 moles of O₂x = \frac{3 \times 3.80}{2} = 5.70 \text{ moles of O₂}x=23×3.80=5.70 moles of O₂
Step 3: Find the mass of O₂ required
The molar mass of O₂ is given as 32.0 g/mol. So:Mass of O₂=5.70 moles×32.0 g/mol=182.4 grams\text{Mass of O₂} = 5.70 \text{ moles} \times 32.0 \text{ g/mol} = 182.4 \text{ grams}Mass of O₂=5.70 moles×32.0 g/mol=182.4 grams
Rounded to three significant figures, the mass is 182 grams, which matches option D.
Explanation:
This problem involves stoichiometric calculations based on a balanced chemical reaction. The reaction provided shows how aluminum reacts with oxygen to form aluminum oxide. The key to solving this is understanding the mole ratios from the balanced equation.
In this equation, 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide. The coefficients in the equation indicate the proportions in which substances react and are produced. Using these ratios, one can determine how much of one reactant is needed to produce a given amount of product.
We are asked to find the mass of oxygen gas needed to produce 3.80 moles of aluminum oxide. From the equation, every 2 moles of aluminum oxide require 3 moles of oxygen gas. Using simple cross multiplication, we calculated that 3.80 moles of aluminum oxide need 5.70 moles of oxygen gas.
Next, using the molar mass of oxygen gas (32.0 g/mol), we convert moles of oxygen to grams. The calculation shows that 5.70 moles of oxygen corresponds to 182.4 grams, which rounds to 182 grams. This matches option D.
Stoichiometry like this is essential for predicting the amounts of substances needed or produced in chemical reactions, allowing chemists to plan and control reactions efficiently.
