Unit #1Unit #2Unit #3Unit #4Unit #5Unit #6Unit #7Unit #8Unit #9Unit #10
spacerUNIT # 8  
spacerspacerIntroduction
spacerspacerObjectives
spacerspacerReading


spacer
8.1 Conventions
spacerand Symbols
spacer8.2 Balancing
spacerChemical
spacerEquations

spacer8.3 Methods of
spacerBalancing
spacerEquations

spacerBalance by Inspection
spacerBalance algebraically
spacer8.4 Types of
spacerChemical
spacerReactions
spacer8.5 Calculations
spacerBased on
spacerChemical
spacerEquations
spacerMole-mole calculations
spacerMass-mass calculations
spacerMass-mole calculations
spacerLimiting Reagent calculations



spacerProblems
spacer1 | 2
| 3 | 4

Unit #8 CHEMICAL REACTIONS

8.5 - Calculations based on Chemical Equations

In section 8.2, we learned that in balancing chemical equations, coefficients can be inserted into the chemical equations so that

the number of atoms of each type on the LEFT-HAND-SIDE = the number of atoms of each type on the RIGHT-HAND-SIDE

With these coefficients, we can write a ratio of any two or more substances in the chemical equation.

Let's take a look at the balanced chemical equation for the synthesis of water:

2 H2 (g) + O2 (g) reaction arrow 2 H20 (l) *

* The coefficient in front of O2 (g) is by default "1". Coefficient of "1" is usually not written in.

Here are some ratios that we can write from the above balanced chemical equation:

  1. 2 molecules of H2 (g) requires 1 molecule of O2 (g) to produce 2 molecules of H20 (l).

    The important thing is, the ratio is 2:1:2 for the substances H2 : O2 : H20 .

    In any combination of any two substances, the coefficients in the balanced chemical equation tells us: the ratio between H2 : O2 is always 2:1.
    the ratio between H2 : H20 is always 2:2, or 1:1.
    the ratio between O2 : H20 is always 1:2.

  2. If we scale up 2 times, then:
    spacer4 molecules of H2 (g) requires 2 molecules of O2 (g) to produce 4 molecules of H20 (l).

    This ratio of 2:1:2 is always maintained no matter how much we scale up.
  1. If we scale up 12 times, then:
    spacer2 dozens molecules of H2 (g) requires 1 dozens molecules of O2 (g) to produce 2 dozens molecules of H20 (l).

    This ratio of 2:1:2 is always maintained no matter how much we scale up.

  2. If we scale up to a mole, then:
    spacer2
    moles of H2 (g) requires 1 mole of O2 (g) to produce 2 moles of H20 (l).

    This ratio of 2:1:2 is always maintained no matter how much we scale up.

    I repeat:

    The important thing is, the ratio is always 2:1:2 for the substances H2 : O2 : H20 .

    The coefficients give the mole ratio of the reactants and products.

    In any combination of any two substances, the coefficients in the balanced chemical equation tells us: the ratio between H2 : O2 is always 2:1.
    the ratio between H2 : H20 is always 2:2, or 1:1.
    the ratio between O2 : H20 is always 1:2.

This is known as stoichiometric ratio. In unit 7, we learned that moles and grams can be interconverted by a compound's molar mass (section 7.5.1). By knowing the mole ratio of the substances in the reaction, it enables us to calculate the mass of the reactants consumed and the mass of products formed.

We must master calculations involving the different types of stoichiometric problems that arise from a balanced chemical equation. They are:

  1. Mole-mole Calculations (section 8.5.1)
  2. Mass-mass Calculations (section 8.5.2)
  3. Mass-mole Calculations (section 8.5.3)
  4. Excess reactant and limiting reagent problems (section 8.5.4)

When you are going through these sections, you need to go slowly and thoroughly to understand the logic in each step. Refer to your textbook for more examples of each type of stoichiometric problems.

book iconSection 10.1
spacerInformation Obtained from a Balanced Chemical Equation ..p249


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