Balanced equation: 3Cu(SO4) + 2Al -gt; 2Al(SO4)3 + 3CuBeaker A: Mass of aluminum foil was 0.11g and moles of CuSO4 was 0.02mol.Beaker B: Mass of aluminum foil was 0.13g and moles of CuSO4 was 0.01molPlease explain howThanks so much!
Look at the periodic table and count from the first column on the left
Energy Levels: period Valence e-s: group
Pisgahchemist gave you a reply earlierBut I guess you didn't care for his explanationHere's mineIn the activity series Aluminum is above CopperThat is, in your solution of CuSO4 you have Cu^2+ ions (and SO4^2-)When you add Aluminum foil to the solution (Al), the aluminum will transfer electrons to the Cu^2+ ions, and aluminum ions will result (Al^3+) 2 Al + 3Cu^2+ - 2 Al^3+ + 3Cu Beaker A: moles Al 0.11g x 1mol/27g 0.004 moles moles Cu^2+ 0.02 moles From balanced equation, mole ratio of Cu^2+ to Al is 3 : 2For 0.004 moles Al, you need 3/2 x 0.004 moles Cu^+2 0.006 moles Cu^2+There are 0.02 moles of Cu^2+ so Al is limitingBeaker B: moles Al 0.13g x 1mol/27g 0.005 moles moles Cu^2+ - 0.01 moles For 0.005 moles Al you need 3/2 x 0.005 moles Cu^2+ 0.0075 moles Cu^2+There are 0.01 moles of Cu^2+ so Al is limiting.
