% Yield

[soby10]

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Why is 815g the correcr answer... isn't it actual/theor X100% so i got 727g as theor..plugged into equation..
89.3%=(x/727g)X100% i got 64.92g as theor. yield...how is 815g correct?Thanks

In the Hall-Heroult process, anhydrous aluminum oxide is reduced to aluminum according to the chemical equation:

2 Al2O3 (s) ---> 4 Al (l) + 3 O2 (g)

Calculate the mass of aluminum oxide reacted if 385 g of Al was produced and the percent yield was found to be 89.3%. 814.6 g215.6 g1630 gCorrect answer:815 g727 g

 

[joshinjosh]

here's how i did it.

look for mass percent of Al2O3. Al2=27*2=54 and O3=16*3=48. 54 and 48 is roughly half and half. So aluminum oxide by mass is half aluminum, half oxygen.

stoiciometrically, in the equation there are 4 Al on the left and 4 Al on the right.

you obtain 385g of Al. that means 385g of the aluminum in Aluminum oxide is 385g. since we know that aluminum in aluminum oxide by weight is 50%, aluminum oxide must be 2*385g = 770g.

770g of al2o3 is for 100% yield. but experimentally, we got ~90%. that means experimentally, you lost ~10%. so add ~10 to 770g, and you get around 840g! 815g is closest by far! that is your answer

hope this helps

 

[Geekchick921]

Here's how I did it.

Firstly, if 385g of Al is 89.3% yield, then ideally, this reaction should have produced 431g (385/.893). It's just stoic after that.

431g Al x (1 mole Al/27g Al) x (2 moles Al2O3/4 moles Al) x (102g Al2O3/1 mole Al2O3) = 814g Al2O3. Close enough to 815 (depends on how much you round).

 

[unique135]

Why is 815g the correcr answer... isn't it actual/theor X100% so i got 727g as theor..plugged into equation..
89.3%=(x/727g)X100% i got 64.92g as theor. yield...how is 815g correct?Thanks

In the Hall-Heroult process, anhydrous aluminum oxide is reduced to aluminum according to the chemical equation:

2 Al2O3 (s) ---> 4 Al (l) + 3 O2 (g)

Calculate the mass of aluminum oxide reacted if 385 g of Al was produced and the percent yield was found to be 89.3%. 814.6 g215.6 g1630 gCorrect answer:815 g727 g

Where did you get this problem from? Is it from MCAT prep books?



This is how i would have done...but I like the way Joshinjosh did.

89.3 is ~90%

Step 1: 385/0.90 = ~430g would 100% yield. (not much calculation here)

Step 2: 430g/27 = ~16 moles of Al (little calculation here)

Step 3: 16 moles of Al must be equal 8 moles of Al2O3 (Mental)

Step 4: 8 moles x 102 gram/mole = 816g (Mental)