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7 April, 20:57

What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?

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  1. 7 April, 21:18
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    3 H2SO4 + 2 Al (OH) 3 → Al2 (SO4) 3 + 6 H2O

    (2.14 g Al (OH) 3) / (78.0036 g Al (OH) 3/mol) x (3 mol H2SO4 / 2 mol Al (OH) 3) / (0.210 mol/L H2SO4) =

    0.19596 L = 196 mL H2SO4
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