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4.2: Chemical Equations and Stoichiometry

  • Page ID
    36989
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    \[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{4.2.1}\]

    Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?

    The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined in , can be adapted as follows:

    1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:

    \[ moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \]

    2. According to the balanced chemical equation, 6 mol of CO2 is produced per mole of glucose; the mole ratio of CO2 to glucose is therefore 6:1. The number of moles of CO2 produced is thus

    \[ moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \]

    \[ = 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \]

    \[ = 1.51 \, mol \, CO_2 \]

    3. Use the molar mass of CO2 (44.010 g/mol) to calculate the mass of CO2 corresponding to 1.51 mol of CO2:

    \[ mass\, of\, CO_2 = 1.51 \, mol \, CO_2 \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.5 \, g \, CO_2 \]

    \[ 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \]

    Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (For more information about rounding and significant digits, see Essential Skills 1 in , .) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced.

    The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example 11, the mass of one reactant that is required to consume a given mass of another reactant.

    Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.

      Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): https://youtu.be/74mHV0CZcjw


    4.2: Chemical Equations and Stoichiometry is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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