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How many moles of KOH are measured in thisexperiment? See the volume and molarity information given. 2.How many moles of KOH reacted with the amount of Al you used? Refer to the reaction equation in the first few lines of the experiment.3.How many moles of KOH were left over?

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Answer:

0.022 moles of potassium hydroxide reacted with Aluminum and 0.013 moles of potassium hydroxide were left over.

Explanation:

2Al(s) + 2KOH(aq) + 6H2O(l) → 2KAl(OH)4(aq) + 3H2(g)

First we obtain the amount in moles of aluminum added in line with the mass of aluminum to be added in the experimental procedure. This shows us that aluminum metal is the limiting reactant.

Secondly we obtain the amount of KOH in 25mL of 1.4M KOH solution(see attached image). This gives the amount of KOH availiable for reaction.

2 moles of Aluminum metal reacts with 2 moles of KOH according to the balanced reaction equation. Hence 0.022 moles of aluminum will react with 0.022 moles of KOH.

The amount of KOH left unreacted is 0.035-0.022= 0.013 moles of KOH.

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