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22 May, 16:05

How many grams of NaOH needed to completely neutralize 3L of 1.75M HCL

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  1. 22 May, 18:57
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    209.98 g of NaOH

    Explanation:

    We are given;

    Volume of HCl as 3 L Molarity of HCl as 1.75 M

    We are required to calculate the mass of NaOH required to completely neutralize the acid given.

    First, we write a balanced equation for the reaction between NaOH and HCl

    That is;

    NaOH + HCl → NaCl + H₂O

    Second, we determine the number of moles of HCl

    Number of moles = Molarity * Volume

    = 1.75 M * 3 L

    = 5.25 moles

    Third, we use the mole ratio to determine the moles of NaOH

    From the reaction,

    1 mole of NaOH reacts with 1 mole of HCl

    Therefore;

    Moles of NaOH = Moles of HCl

    = 5.25 moles

    Fourth, we determine the mass of NaOH

    Molar mass of NaOH = 39.997 g/mol

    Mass of NaOH = 5.25 moles * 39.997 g/mol

    = 209.98 g

    Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl
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