Verified answer

1.

Considering the equation, we find that 1 mole of Al2O3 needs atleast 3 moles of Cu to form 2 moles of Al and 3 moles of CuO.

Since we have only 1 mole of Al2O3 but an unlimitted amuount of Cu, we can say that Al2O3 is the limitting reagent (the substance which limits the amount of products formed a chemical reaction).

To react with 1 mole of Al2O3, we actually need only 3 moles of Cu, but they have provided us with plenty of it. Dosen't matter, we don't need the extra amount of Cu that they have given us.

Therefore(Hope u have already guessed the answer out!)

We can produce only 2 moles of Al with the given amount of reactants.

2.

The molar mass of Al2O3 is = 2*26.9+ 3*16

= 101.8(approximately)

Therefore, if we have 408g of Al2O3,

(We know that- No. of moles of a substance = Given mass of the substance/ Molar mass of the substance -----------equation 1)

Applying in the formula, we find that

No of moles= 408/101.8

= 4.007

= 4 moles

Therefore, 408g of Al2O3 is equivalent to having 4 moles of Al2O3.

From the equation, we know that 1 mole Al2O3 needs 3 Mole Cu to produce 2 Mole Al

Sice the amount of Cu given is unlimitted, we don't need to consider it.

We just need this part of it-

ie. 1 mole Al2O3 gives 2 Mole Al

Therefore, 4 mole Al2O3 should give (2*4)mole Al

= 8 mole Al

Therefore 408 grams of Al2O3 gives us 8 mole Al.

Now using the equation(1) [ scroll a bit up to see it again]

7.6= Mass given(in grams)/ Molar mass of Al

Molar mass of Al is 26.9 ==>27

Therefore, the Mass given (in grams) = 8*27

= 216 grams

Threfore, the final conclusion is that

408g Al2O3 produces 216g of Al with an unlimitted supply of Cu.

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CAUTION!!!

IN THE ABOVE 2 PROBLEMS, THE AMOUNT OF Cu IS NOT BEING CONSIDERED BECAUSE IT IS PROVIDED IN UNLIMITTEDLY. IF THE AMOUNT OF Cu WERE LIMITTED, WE WOULD HAVE TO CONSIDER IT TOO. THEN THERE WOULD BE A LITTLE MORE CALCULATIONS INITIALLY, BUT REST ASSURED THAT THE HEART OF THE SOLVING METHOD WOULD REMAIN LARGELY UNAFFECTED THOUGH THE NUMBERS AND YOUR ANSWER MIGHT CHANGE A BIT.

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Hope I helped!

1.Consider the balanced equation Al(2)O(3) + 3 Cu -> 2Al +3CuO. If you have 1 mole of Al(2)O(3) and an unlimited amount of Cu, how many moles of Al can you produce?

1 mol Al2O3 (2 mol Al / 1 mol Al2O3) = 2 moles Al

2. If you have 408 grams of Al(2) O(3) and an unlimited supply of Cu, how many grams of Al can you produce (round to the nearest gram)? Show calculations.

molar mass Al2O3 = 2(26.98) + 3(16) = 101.96g

molar mass Al = 26.98g

408g Al2O3 (1 mol Al2O3 / 101.96g Al203) (2 mol Al / 1 mol Al2O3) (26.98g Al / 1 mol Al) = 216g Al

Bill