Assuming you start with 1.25 g of pure aluminum, calculate the following:
1) the amount of potassium hydroxide, koh, in grams needed to react with all of the 1.25 g of al. the reaction is:
2al(s) + 2 koh(aq) + 6h2o(l) - 2k+ (aq) + 2 al(oh)4- (aq) + 3h.

The amount of potassium hydroxide (KOH) needed to react with all of the 1.25 g of Aluminum is 2.58 g.

The given parameters;

The chemical reaction of the given elements is given as;

From the reaction above we can deduce the following;

2 moles of AL ------------- 2 moles of KOH

How many moles of 1.25 g of aluminum will be required ?

The mass of KOH required for 0.046 mole of Aluminum is calculate as;

mass required = 0.046 x 56

mass required = 2.58 g

Thus, the amount of potassium hydroxide (KOH) needed to react with all of the 1.25 g of Aluminum is 2.58 g.

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a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.