how many mL of a 4 M AI(OH)3 solution are needed to react with 125 mL of 2 M HCI solution? AI(OH)3 (s) + 3HCI (aq) (Arrow>) AlCI3(aq)+3 H2O (l)

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how many mL of a 4 M AI(OH)3 solution are needed to react with 125 mL of 2 M HCI solution?

AI(OH)3 (s) + 3HCI (aq) (Arrow>) AlCI3(aq)+3 H2O (l)

 

Expert Solution
Step 1

AI(OH)3 (s) reacts with HCI (aq) to form AlCI3(aq) and H2O (l). The balanced chemical reaction is as follows:

AI(OH)3 (s) + 3HCI (aq) (Arrow>) AlCI3(aq)+3 H2O (l)

The stoichiometry of the balanced chemical reaction states that one mole of AI(OH)3 (s) reacts with three moles of HCI (aq).

Step 2

At the endpoint we can say that:

3×molarity of Al(OH)3×volume of Al(OH)3=molarity of HCl×volume of HCl

We know that the molarity of Al(OH)3 is = 4M

the molarity of HCl is = 2M

the volume of HCl is = 125 mL

The required volume of Al(OH)3 is calculated as follows:

volume of Al(OH)3=molarity of HCl×volume of HCl3×molarity of Al(OH)3volume of Al(OH)3=2 M×125 mL3×4Mvolume of Al(OH)3=2 ×125 mL3×4 volume of Al(OH)3=20.8 Mvolume of Al(OH)3=21 mL

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