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31 March, 18:25

How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda? Show your work.

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  1. 31 March, 21:11
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    75mL

    Explanation:

    Step 1:

    The balanced equation for the reaction. This is illustrated below:

    C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3CO2 + 3H2O

    Step 2:

    Conversion of 15g of baking soda (NaHCO3) to mole.

    This is illustrated below:

    Mass of NaHCO3 = 15g

    Molar Mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 23 + 1 + 12 + 48 = 84g/mol

    Number of mole = Mass/Molar Mass

    Number of mole of NaHCO3 = 15/84 = 0.179 mole

    Step 3:

    Determination of the number of mole of citric acid (C6H8O7) produced by the reaction. This is illustrated below:

    From the balanced equation above,

    1 mole of C6H8O7 reacted with 3 moles of NaHCO3.

    Therefore, Xmol of C6H8O7 will react with 0.179 mole of NaHCO3 i. e

    Xmol of C6H8O7 = 0.179/3

    Xmol of C6H8O7 = 0.06 mole.

    Step 4:

    Determination of the volume of C6H8O7 needed for the reaction. This is illustrated below:

    Molarity of C6H8O7 = 0.8 M

    Mole of C6H8O7 = 0.06

    Volume = ... ?

    Molarity = mole / Volume

    Volume = mole / Molarity

    Volume = 0.06/0.8

    Volume = 0.075L

    Converting 0.075L to mL, we have:

    1L = 1000mL

    Therefore 0.075L = 0.075 x 1000 = 75mL.

    Therefore, 75mL of citric acid (C6H8O7) is needed for the reaction.
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