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Lowell High School Chemistry

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Practice Problem #3:

How many grams of aluminum sulfate are produced if 23.33 g Al reacts with 74.44 g CuSO4?

Al (s)  +  CuSO4 (aq)  à  Al2(SO4)3 (aq)  +  Cu (s)  [unbalanced]

Answer to Practice Problem #3:

1. Balance your equation first.

2Al (s)  +  3CuSO4 (aq)  à  Al2(SO4)3 (aq)  +  3Cu (s)                    [balanced]

2. Convert from grams to moles using molar mass.

To get moles from grams of aluminum (Al):   [molar mass = 27 g/mol]

23.33 g Al X  ( 1 mol  /  27 g ) = 0.8640741 mol Al

To get moles from grams of copper (II) sulfate (CuSO4):       [molar mass = 159.612 g/mol]

74.44 g CuSO4  X  ( 1 mol  / 159.612 g ) = 0.466381 mol CuSO4

3. Determine the limiting reagent [if necessary] (Use mole ratios to figure out.)

Figure out how many moles of CuSO4 would need to react with 0.8640741 mol Al:

* From your balanced equation, the ratio is 3 mol CuSO4 : 2 mol Al.

0.8640741 mol Al  X  ( 3 mol CuSO4 / 2 mol Al )  = 1.296 mol CuSO4 needed

* But you have 0.466381 mol CuSO4 to start with and need 1.296 mol CuSO4 to react, so you have not enough CuSO4, which makes CuSO4 the limiting reagent.

OR

Figure out how many moles of Al would need to react with 0.466381 moles CuSO4:

* From your balanced equation, the ratio is 2 mol Al : 3 mol CuSO4.

0.466381 mol CuSO4  X  ( 2 mol Al / 3 mol CuSO4 )  = 0.3109 mol Al needed

* But you have 0.8641 mol Al to start with and need 0.3109 mol Al to react, so you have lots of Al in excess, which makes CuSO4 the limiting reagent.

4. Use ratios to find the moles of the reactant or product you need to find.

Since CuSO4 is the limiting reagent. You use the moles of CuSO4 if you have to calculate how many moles of aluminum sulfate (the product you’re trying to find) are being produced. Use your mole ratios from your balanced equation:

0.466381 mol CuSO4 X  ( 1 mol Al2(SO4)3 / 3 mol CuSO4 )  = 0.1555 mol Al2(SO4)3 produced

5. Convert from moles back to grams of the new substance using that substance’s molar mass.

To get grams from moles of aluminum sulfate (Al2(SO4)3):    [molar mass = 342.3 g/mol]

0.1555 mol Al2(SO4)3  X  ( 342.3 g  /  1 mol ) = 53.23 g Al2(SO4)3 produced

Answer to Problem: 53.23 g Al2(SO4)3 produced

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