**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 CuSO_{4}?

Al (s) + CuSO_{4} (aq) à Al_{2}(SO_{4})_{3} (aq) + Cu (s) [unbalanced]

**Answer to Practice Problem #3:**

**1.** **Balance your equation first.**

**2**Al (s) + **3**CuSO_{4} (aq) à Al_{2}(SO_{4})_{3} (aq) + **3**Cu (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 (CuSO_{4}): [molar mass = 159.612 g/mol]

74.44 g CuSO_{4} X ( 1 mol / 159.612 g ) = 0.466381 mol CuSO_{4}

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

Figure out how many moles of CuSO_{4} would need to react with 0.8640741 mol Al:

* From your balanced equation, the ratio is 3 mol CuSO_{4} : 2 mol Al.

0.8640741 mol Al X ( 3 mol CuSO_{4} / 2 mol Al ) = 1.296 mol CuSO_{4} needed

* But you have 0.466381 mol CuSO_{4} to start with and need 1.296 mol CuSO_{4} to react, so you have not enough CuSO_{4}, which makes CuSO_{4} the limiting reagent.

OR

Figure out how many moles of Al would need to react with 0.466381 moles CuSO_{4}:

* From your balanced equation, the ratio is 2 mol Al : 3 mol CuSO_{4}.

0.466381 mol CuSO_{4} X ( 2 mol Al / 3 mol CuSO_{4} ) = 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 CuSO_{4} the limiting reagent.

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

Since CuSO_{4} is the limiting reagent. You use the moles of CuSO_{4} 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 CuSO_{4} X ( 1 mol Al_{2}(SO_{4})_{3} / 3 mol CuSO_{4} ) = 0.1555 mol Al_{2}(SO_{4})_{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 (Al_{2}(SO_{4})_{3}): [molar mass = 342.3 g/mol]

0.1555 mol Al_{2}(SO_{4})_{3} X ( 342.3 g / 1 mol ) = **53.23 g Al _{2}(SO_{4})_{3} produced**

**Answer to Problem:** **53.23 g Al _{2}(SO_{4})_{3} produced**

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