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Masja [62]
1 year ago
10

6.0 mol Al reacts with 4.0 mol O2 to form Al2O3.

Chemistry
1 answer:
Tema [17]1 year ago
6 0

Answer:

3.0 moles Al₂O₃

Explanation:

We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.

4 Al + 3 O₂ -----> 2 Al₂O₃

6.0 moles Al            2 moles Al₂O₃
----------------------  x  -------------------------  =  3.0 moles Al₂O₃
                                    4 moles Al

4.0 moles O₂           2 moles Al₂O₃
----------------------  x  -------------------------  =  2.7 moles Al₂O₃
                                    3 moles O₂

As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.

However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.

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Answer:

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Explanation:

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If 2.5L of solution is diluted to prepare 1.7L of a 0.8M solution, what was the original concentration?
mamaluj [8]

Answer:

0.544 M

Explanation:

First find the moles in the final solution

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Answer:

Answer is given below.

Explanation:

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So anode is Cu and cathode is Ag.

oxidation: Cu-2e^{-}\rightarrow Cu^{2+}(aq.)

[reduction: Ag^{+}+e^{-}\rightarrow Ag]\times 2

-----------------------------------------------------------------------------------------------

chemical equation: Cu+2Ag^{+}(aq.)\rightarrow Cu^{2+}(aq.)+2Ag

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Reducing agent is that species which gives electron to another species. Here Cu gives electron to Ag^{+}(aq.) . Hence Cu  is the reducing agent.

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Answer:

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