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How many mL of 1.555 M NaOH are required to completely neutralize 3.587 g of H2C2O4 (MM = 90.0) ?

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  1. 8 June, 00:17
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    51.45 mL

    Explanation:

    We are given;

    Molarity of NaOH as 1.555 M Mass of the acid, H₂C₂O₄ as 3.587 g

    We are required to determine the volume of NaOH required.

    Step 1: Write the balanced equation for the reaction.

    The balanced equation for the reaction between NaOH and H₂C₂O₄ will be;

    2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

    Step 2: Determine the number of moles of the acid, H₂C₂O₄

    We know that, moles are calculated by dividing the mass of the compound by its molar mass.

    Moles = Mass : Molar mass

    Therefore;

    Moles of H₂C₂O₄ = 3.587 g : 90.0 g/mol

    = 0.040 moles

    Step 3: Determine the number of moles of NaOH that reacted

    From the equation, the mole ratio of NaOH to H₂C₂O₄ is 2 : 1

    Therefore; Moles of NaOH = Moles of H₂C₂O₄ * 2

    Thus;

    Moles of NaOH = 0.040 moles * 2

    = 0.080 moles

    Step 4: Determine the volume of NaOH that reacted

    Molarity of NaOH = 1.555 M

    But; Molarity = Moles : Volume

    Rearranging the formula;

    Volume = Moles : Molarity

    Therefore;

    Volume of NaOH = 0.080 moles : 1.555 M

    = 0.05145 L

    But, 1 L = 1000 mL

    Therefore;

    Volume = 51.45 mL

    Hence, the volume of HCl required is 51.45 mL
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