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26 November, 00:12

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water. What is the theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 4.6 g of sodium hydroxide?

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Answers (2)
  1. 26 November, 00:15
    0
    5.28g

    Explanation:

    The equation for the reaction between is given below:

    HCl + NaOH - > NaCl + H2O

    Let us determine which will be the limiting reactant.

    Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

    Mass of HCl = 3.3g

    Number of mole = Mass / Molar Mass

    Number of mole of HCl = 3.3/36.5 = 0.09mole

    Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

    Mass of NaOH = 4.6g

    Number of mole = Mass / Molar Mass

    Number of mole of NaOH = 4.6/40 = 0.115mole

    From the equation,

    1mole of HCl required 1mole of NaOH.

    Therefore, 0.09mole of HCl will also require 0.09mole of NaOH.

    Now, we see that NaOH is excess and HCl is limiting.

    Now we can calculate the theoretical yield as follows:

    Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

    From the equation,

    36.5g of HCl produced 58.5g of NaCl.

    Therefore, 3.3g of HCl will produce = (3.3 x 58.5) / 36.5 = 5.28g

    Therefore, the theoretical yield of NaCl is 5.28g
  2. 26 November, 00:41
    0
    The theoretical yield of sodium chloride is 5.3 grams

    Explanation:

    Step 1: Data given

    Mass of hydrochloric acid (HCl) = 3.3 grams

    Mass of sodium hydroxide (NaOH) = 4.6 grams

    Molar mass HCl = 36.46 g/mol

    Molar mass NaOH = 40.0 g/mol

    Step 2: The balanced equation

    HCl + NaOH → NaCl + H2O

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles HCl = 3.3 grams / 36.46 g/mol

    Moles HCl = 0.0905 moles

    Moles NaOH = 4.6 grams / 40.0 g/mol

    Moles NaOH = 0.115 moles

    Step 4: Calculate limiting reactant

    For 1 mol HCl we need 1 mol NaOH to produce1 mol NaCl and 1 mol H2O

    HCl is the limiting reactant. It will completely be consumed (0.0905 moles). NaOh is in excess. There will react 0.0905 moles. There will remain 0.115 - 0.0905 = 0.0245 moles NaOH

    Step 5: Calculate moles NaCl

    For 1 mol HCl we need 1 mol NaOH to produce1 mol NaCl and 1 mol H2O

    For0.0905 moles HCl we'll have 0.0905 moles NaCl

    Step 6: Calculate the theoretical yield of NaCl

    Mass NaCl = moles NaCl * molar mass NaCl

    Mass NaCl = 0.0905 * 58.44 g/mol

    Mass NaCl = 5.3 grams

    The theoretical yield of sodium chloride is 5.3 grams
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