How many mL of .200M HCl are needed to completely neutralize 3.50g of Al(OH)3?
Determine moles of Al(OH)3 n(Al(OH)3) = 3.50 g / 78.0 g/mol = 4.49x10^-2 mol Determine moles of HCl reacted n(HCl) = 3 x n(Al(OH)3) = (3)(4.49x10^-2) = 0.135 mol Determine volume of HCl required using equation for molarity Molarity = moles / volume volume = moles / molarity = 0.135 mol / 0.200 M = 0.675 L = 675 mL
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