chemistry help, neutralizing

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

Apr 22nd, 2015

3HCl + Al(OH)3   = AlCl3   + 3H2O

molecular mass of Al(OH)3 = 78 g/mol

Hence number of moles of Al(OH)3 

= wt of Al(OH)3 / molar mass of Al(OH)3

= 3.5/78

=0.0448 moles

Now for 1 moles of Al(OH)3, number of moles of HCl required for neutralizing = 3

Hence for 0.0448 moles of Al(OH)3, number of moles of HCl requirred  = 3*0.0448 = 0.1344 moles of HCl

Now number of moles = molarity *volume

Hence volume of HCL required 

= 0.200 (M)* 0.1344*( moles)

=0.02688 L

=26.88 ml

Hence volume of HCl required = 26.88 ml


Apr 22nd, 2015
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Apr 22nd, 2015

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