How do you solve a stoichiometry problem?

Answer 1

You use a series of conversion factors to get from the units of the given substance to the units of the wanted substance.

There are four steps in solving a stoichiometry problem:

  1. Write the balanced chemical equation.
  2. Convert the units of the given substance (A) to moles.
  3. Use the mole ratio to calculate the moles of wanted substance (B).
  4. Convert moles of the wanted substance to the desired units.

    The flow chart below summarizes the process.


    (From MillingsChem)

    NOTE: The mole ratio of A to B is central to all the calculations.

    EXAMPLE:

    What mass of chlorine does the decomposition of 64.0 g of AuCl₃ produce?

    Solution:

    1. Write the balanced chemical equation.

    #"2AuCl"_3 → "2Au" + "3Cl"_2#

    2. Convert grams of #"AuCl"_3# to moles of #"AuCl"_3#.

    #64.0 color(red)(cancel(color(black)("g AuCl"_3))) × ("1 mol AuCl"_3)/(303.3 color(red)(cancel(color(black)("g AuCl"_3)))) = "0.211 mol AuCl"_3#

    3. Use the molar ratio to convert moles of #"AuCl"_3# to moles of #"Cl"_2#.

    #0.211 color(red)(cancel(color(black)("mol AuCl"_3))) × ("3 mol Cl"_2)/(2 color(red)(cancel(color(black)("mol AuCl"_3)))) = "0.317 mol Cl"_2#

    4. Convert moles of #"Cl"_2# to grams of #"Cl"_2#.

    #0.317 color(red)(cancel(color(black)("mol Cl"_2))) × ("70.91 g Cl"_2)/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "22.4 g Cl"_2#

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Answer 2
  1. Balance the chemical equation.
  2. Convert given quantity to moles.
  3. Use mole ratios from the balanced equation to find moles of the desired substance.
  4. Convert moles to the desired units (grams, liters, etc.) using appropriate conversion factors.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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