Theoretical yield of Al₂O₃: 1.50 mol.
<h3>Explanation</h3>
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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.
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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.

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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Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
The letter only has reflection symmetry, if you rotated it it would look the same.
Answer: potassium nitrate
hope this helps