solid aluminum and oxygengas react to form solid aluminum oxide. suppose you have 7.0 mol of al and 11.0 mol of o2 in a reactor. calculate the largest amount of that could be produced. round your answer to the nearest .

Respuesta :

The largest amount of Al₂O₃ produced from the reaction is 3.5 moles

What are limiting reactants?

A reactant that is completely consumed at the end of a chemical reaction is known as the limiting reagent. This reagent regulates the amount of product produced because the reaction cannot proceed without it.

Finding the limiting reactant,

4Al + 3O₂ —> 2Al₂O₃

From the balanced equation above,

4 moles of Al reacted with 3 moles of O₂.

Therefore,

7 moles of Al will react with = (7 × 3)/4 = 5.25 moles of O₂

From above, only 5.25 moles of O₂ out of 11 moles given, is needed to completely react with 7 moles of Al.

Therefore, Al is the limiting reactant and O₂ is the excess reactant.

Now, the largest amount of Al₂O₃ produced from the reaction. This can be obtained by using the limiting reactant as shown below

From the balanced equation above,

4 moles of Al reacted to produce 2 moles of Al₂O₃.

Therefore,

7 moles of Al will react to produce = (7 × 2)/4 = 3.5 moles of Al₂O₃.

Thus, the largest amount of Al₂O₃ produced from the reaction is 3.5 moles

To learn more about limiting reactants follow the link:

brainly.com/question/15230635

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