how many grams of aluminum hydroxide are needed to completely react with 75.0 mL of 0.175 M chloric acid

Question

please help with this question

Answer 1

First write the balanced equation for the reaction:

Al(OH)₃ + 3HClO ==> 3H₂O + Al(ClO)₃

Next, find the moles of HClO present:

75.0 ml x 0.175 mmoles/ml = 13.125 mmoles = 0.0131 moles

Next, find moles Al(OH)₃ needed to neutralize 0.0131 moles HClO:

0.0131 moles HClO x 1 mole Al(OH)₃/3 moles HClO = 0.004375 moles Al(OH)₃ needed

Finally, convert moles to grams:

0.004375 moles Al(OH)₃ x 78.0 g/mole = 0.341 g (to 3 sig. figs.)