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The mass of KOH reacted is 2.576 g. To have a 10% excess KOH, i need 2.833 g of KOH.

The equation of the reaction is;

2 Al(s) + 2 KOH(aq) + 6H20 (l) --------> 2K+ (aq) + 2 Al(OH)4^- (aq) + 3H2(g)

We are told that 1.25 g of Al was used in the reaction, hence;

Number of moles of Al = 1.25 g/27 g/mol = 0.046 moles

To obtain the amount of KOH required for the reaction;

2 moles of Al reacts with 2 moles of KOH

0.046 moles of Al also reacts with 0.046 moles of KOH

Mass of KOH reacted = 0.046 moles of KOH × 56 g/mol = 2.576 g

To have a 10% excess KOH;

Let the needed mass of KOH be x

x =2.576 g + (0.1 × 2.576 g)

x = 2.833 g

Learn more: https://brainly.com/question/10079361

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: 2 Al(s) + 2 KOH(aq) + 6H20 (1) --->  2K+ (aq) + 2 Al(OH)4^- (aq) + 3H2(g)  
  • 2) The amount of KOH needed, in grams, in reaction 1 if you want to have 10% excess KOH.

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