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Lena [83]
3 years ago
15

What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
4 0

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3;

4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}.

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}.

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

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We can solve this problem by keeping in mind<em> Dalton's law of partial pressures</em>, which states that the total pressure of a mixture of gases is equal to the sum of each gas' partial pressures. In other words, for this case:

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