How many grams of aluminum nitrate will be formed when 167 mL of 2.50M of nitric acid are combined with an excess of aluminum hydroxide according to the following reaction? 3 HNO3 + Al(OH)3 → 3 H2O + Al(NO3)3
Added by Charles T.
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50M solution. Given: Volume of nitric acid = 167 mL = 0.167 L Molarity of nitric acid = 2.50 mol/L Number of moles of HNO3 = Molarity x Volume Number of moles of HNO3 = 2.50 mol/L x 0.167 L Number of moles of HNO3 = 0.4175 mol Show more…
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