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Limited Reactants Help

  1. Jan 19, 2007 #1
    I totally need help on these problems about limited reactants. My teacher gave me an example: You intend to use the following recipe (Caution: it will turn your mouth bright red—be careful):

    1 can frozen white grape juice concentrate

    2 Tbs. red food color

    1 tsp. yellow food color

    Suppose for your movie you need to make 10 batches of this fake blood. In your freezer you have 9 cans of white grape juice, you have 15 tablespoons of red food color and you have 8 teaspoons of yellow food color. Obviously, you need to go to the store, but how many batches can you make?

    9 cans of white grape juice will make 9 batches.

    15 tbs. of red food color can make 7.5 batches (2 tbs. each)

    8 tsp. of yellow food color can make 8 batches.

    So.....The answer must be 7.5 batches if you want to make half a batch.

    Then she gave me these equations to solve:
    1. How many grams SO2 can be formed from 20.0 g of S and 160g O2?
    2. How many gramsSO2 can be formed from 20 g S and 15.0 g of O2?

    (Plese show the balanced equation and your work.)

    The balanced equation is S + O2 ---> SO2. What I don't know how to do is the work to figure out the equations above.
  2. jcsd
  3. Jan 19, 2007 #2
    This is why measurement in moles is so useful. You can come up with certain equivalencies from your balanced equation:

    1 mol S can yield 1 mol SO2
    1 mol O2 can yield 1 mol SO2

    Then convert your given amounts of S and O2 from grams to moles. You can then use those equivalencies. I won't solve your problem but I'll show you how to do a similar problem:

    2H2 + O2 --> 2H2O

    The atomic weight of H is 1 g/mol, so the molecular weight of H2 is 2 g/mol. The atomic weight of O is 16 g/mol, so the molecular weight of O2, is 32 g/mol.

    Now our equivalencies (which we get from the balanced equation above):

    2 mol H2 = 2 mol H2O
    1 mol O2 = 2 mol H2O

    Say we are given 14 g H2 and 50 g O2. Convert these amounts to moles:

    \frac{14\,g\,H_2}{1} \frac{1\,mol\,H_2}{2\,g\,H_2} = 7\,mol\,H_2

    \frac{50\,g\,O_2}{1} \frac{1\,mol\,O_2}{32\,g\,O_2} = 1.5625\,mol\,O_2

    Now use the equivalencies we have from the reaction equation to see how much of the product will be made given our reactants:

    \frac{7\,mol\,H_2}{1} \frac{2\,mol\,H_2 O}{2\,mol\,H_2} = 7\,mol\,H_2 O

    \frac{1.5625\,mol\,O_2}{1} \frac{2\,mol\,H_2 O}{1\,mol\,O_2} = 3.125\,mol\,H_2 O

    As you can see, given our amount of H2 and O2, we can only make 3.125 mol H2O. O2 limits the reaction since it it all used up when this much H2O has been made, production of H2O stops, and we have some extra H2 left over.
    Last edited: Jan 19, 2007
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