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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it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.