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?

Question

Chemistry

Stokes

17 March, 04:33

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)

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Pierce Nelson · Answer 1

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

Trey Myers · Answer 2

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