Find the volume of 0.130 M hydrochloric acid necessary to react completely with 1.55g Al(OH)3.?

Can someone help with this problem, I’ve tried all sorts of equations to figure this out, but cant get the right answer. Thanks!

1 Answer

  • start with balanced equation:

    3 HCl + Al(OH)3 —> AlCl3 + 3 H2O

    MW Al(OH)3 = 78.00344 g/mol so 1.55g / MW = 0.0200 moles aluminum hydroxide

    you need 3 x 0.0200 moles HCl according to the formula.

    0.0600 moles HCl / 0.130 moles/L = 0.46 L = 460 mL

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