How many mL of .200M HCl are needed to completely neutralize 3.50g of Al(OH)3?
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
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)
Hence volume of HCl required = 26.88 ml
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