Mass of beaker Mass of beaker and CuSO4 5H2O Mass of CuSO4·5H20 Mass of Al foil
ID: 1000899 • Letter: M
Question
Mass of beaker Mass of beaker and CuSO4 5H2O Mass of CuSO4·5H20 Mass of Al foil Mass of beaker and copper (first heating) Mass of beaker and copper (second heating) Mass of beaker and copper (third heating, if needed) Mass of copper metal 2.612 g 1:452E Write the balanced reaction of aluminum with copper D) sulfate in aqueous solutions Write the balanced reaction of aluminum with copper aID sulfate in aqueous solution: 20(92) .ix10 mol Moles of copper product based on moles of Al Al: 24.98 8/mo mal Al Moles of copper product based on CuSO 5H-0 43Explanation / Answer
Moles of Cu product based on moles of Al = 0.253 gm*(1 mol Al/26.98 gm Al)*(3 mol Cu/2 mol Al) = 0.0141 mol
Molar mass of CuSO4.5H2O = 63.55 gm + 32.06 gm + 4*(16.00 gm) = 5*(1.008 + 16.00) gm = 244.65 gm/mol (we need to consider the pentahydrate as it is present as water of crystallization with CuSO4)
Moles of Cu product based on CuSO4.5H2O = 11.459 gm*(1 mol CuSO4.5H2O/244.65 gm CuSO4.5H2O)*(3 mol Cu/3 mol CuSO4.5H2O) = 0.0468 mol
The Limiting Reagent is the one that produces the lesser amount of the product; in this case, it is Al (s)
Limiting Reagent: Al
Reactant in excess: CuSO4.5H2O
In order to find out the excess amount of the reactant in excess, we must first calculate how much CuSO4.5H2O has reacted with Al.
As per the stoichiometric equation, 1 mole = 26.98 gm Al reacts with 3 moles = (3*244.65)gm = 733.95 gm CuSO4.5H2O (1:3 molar ratio)
Amount of CuSO4.5H2O reacted with 0.253 gm Al = 0.253 gm Al*(1 mol Al/26.98 gm Al)*(3 mol CuSO4.5H2O/1 mol Al)*(244.65 gm CuSO4.5H2O/1 mol CuSO4.5H2O) = 6.882 gm CuSO4.5H2O
Amount of excess reactant = 11.459 gm – 6.882 gm = 4.577 gm
Amount of excess reactant that hasn’t reacted: 4.577 gm
Next, we need to calculate the theoretical yield of Cu.
Amount of Cu product as per theory = 0.253 gm Al*(1 mole Al/26.98 gm Al)*(3 mol Cu/2 mol Al)*(63.55 gm Cu/1 mol Cu) = 1.788 gm Cu
Theoretical yield of Cu product: 1.788 gm
However, actual amount of Cu product obtained = 0.527 gm
Percentage yield of Cu product = (0.527 gm/1.788 gm)*100 = 29.47
Percentage yield of Cu: 29.47%
A number of reasons can be responsible for not obtaining the desired yield. Some of the reasons are:
1) Human error: Error in weighing of the glassware and/or sample.
2) The reaction is possibly not efficient enough to produce a high yield of Cu. Though this point is more valid in case of organic reaction, but some impurities may have been present in the Al or the CuSO4.5H2O solution used in the reaction. The presence of impurities in the reagents will lower the yield.
3) Contamination of the glassware used in the experiment by impurities can be a possible reason as well.
The balanced equation for the reaction of Cu with atmospheric oxygen is
2Cu (s) + O2 (g) -------> 2 CuO (black)