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25 October, 09:19

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

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  1. 25 October, 12:41
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    74.4 ml of millimeters needed to react with 15 grams of baking soda

    Explanation:

    Given data

    n C8 H8 07 = 0.8 M

    m NaHCO3=15 g

    Solving the problem:

    From the equation we get,

    3x C6 H8 07 = NaHC03

    Finding the molar mass for the baking soda

    NaHCO3 = 23+1+12+3 x16 = 84 g / mol

    so,

    n NaHCO3 = m NaHCO3 / M NaHCO3

    And then,

    n C6 H8 07 = 1/3 * n NaHCO3

    From the definition of the molarity we get

    V C6 H8 O 7=n C6 H8 07/C6 H8 07

    =n NaHCO3/3 * C6 H8 07 =

    m NaHCO3 / M NaHCO3 / 3 * C6 H8 07

    Calculating the equation we get,

    V C6 H8 O 7 = 15/84/3 * 0.8=5/84 * 0.8 = 0.0744L = 74.4 ml

    74.4 ml of millimeters needed to react with 15 grams of baking soda

    n C8 H8 07 = 0.8 M

    m NaHCO3=15 g

    Solving the problem:

    From the equation we get,

    3x C6 H8 07 = NaHC03

    Finding the molar mass for the baking soda

    Nahco3 = 23+1+12+3 x16 = 84 g / mol

    so,

    n NaHCO3 = m NaHCO3 / M NaHCO3

    And then,

    n C6 H8 07 = 1/3 * n NaHCO3

    From the definition of the molarity we get

    V C6 H8 O 7=n C6 H8 07/C6 H8 07

    =n NaHCO3/3 * C6 H8 07 =

    m NaHCO3 / M NaHCO3 / 3 * C6 H8 07

    Calculating the equation we get,

    V C6 H8 O 7 = 15/84/3 * 0.8=5/84 * 0.8 = 0.0744L = 74.4 ml

    74.4 ml of millimeters needed to react with 15 grams of baking soda
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