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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.

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cristisp93

2.58 g of KOH

Explanation:

We have the following chemical reaction:

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

where:

(s) - solid

(aq) - aqueous

(l) - liquid

(g) - gas

number of moles = mass / molecular weight

number of moles of Al = 1.25 / 27 = 0.046 moles

Taking in account the chemical reaction we formulate the following reasoning:

if         2 moles of Al react with 2 moles of KOH

then   0.046 moles of Al react with X moles of KOH

X = (0.046 × 2) / 2 = 0.046 moles of KOH

mass = number of moles × molecular weight

mass of KOH = 0.046 × 56 = 2.58 g

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onyebuchinnaji

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;

  • mass of the pure aluminum, = 1.25 g
  • molecular mass of KOH = 56 g/mol

The chemical reaction of the given elements is given as;

[tex]2Al \ + \ 2KOH \ + 6H_2O \ - 2K + 2Al(OH)^{4-} \ + 3H[/tex]

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 ?

[tex]no \ of \ moles = \frac{1.25}{27} \\\\no \ of \ moles = 0.046[/tex]

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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