chemistry help, neutralizing

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How many mL of .200M HCl are needed to completely neutralize 3.50g of Al(OH)3?

Oct 21st, 2017

3HCl+Al(OH)3---->AlCl3+3H20

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


Please best the solution !!!

Apr 22nd, 2015

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