What volume of 0.200 M aluminum nitrate solution is needed to precipitate 50.0 grams of aluminum hydroxide from an excess of 0.100 M potassium hydroxide solution?

Answer 1

Al(NO₃)₃(aq) + 3KOH(aq) ==> 3KNO₃(aq) + Al(OH)₃(s) ... balanced equation

50.00 g Al(OH)₃ x 1 mol/78 g = 0.744 moles Al(OH)₃ desired

0.744 mol Al(OH)₃ x 1 mol Al(NO₃)₃ / mole Al(OH)₃ = 0.744 moles Al(NO₃)₃ needed

Volume Al(NO₃)₃ needed: (x L)(0.200 mol/L) = 0.744 moles and x = 3.72 L