Question

If you add 35.6mL of 5.20M Hcl to 35.6mL of 7.40M NaOH, the solution that you generate is what percent NaCl? (Keep in mind the following... you must balance the equation to ge the answer and you must also consider what is the final volume of this solution?)(molar mass of NaCl is 58.44g/mol)

Answer 1

A stoichiometrically balanced neutralization reaction between NaOH and HCl is,

NaOH + HCl -----------> NaCl + H₂O.

Molar equivalence says,

1 Mole of NaOH = 1 Mole of HCl = 1 mole of NaCl.

WIth given details let's find number of moles of each reactant involved in reaction,

For NaOH,

M1 = 7.40 M, V1 = 35.6 mL = 0.0356 L hence,

# of moles of NaOH = M1 x V1 = 7.40 x 0.0356 = 0.236

For HCl,

,M2 = 5.20 M, V2 = 35.6 mL = 0.0356 L.

# of moles of HCl = M2 x V2 = 5.20 x 0.0356 = 0.185

As number of moles of HCl is lesser than number of moles NaOH, HCl is limiting reagent.

Hence we write that 0.185 moles of NaOH and 0.185 moles of HCl react to form 0.185 moles of NaCl salt.

# of moles of NaOH unreacted = 0.236 - 0.185 = 0.051.

Let us find out mass of all componants of reaction mixture.

Molar mass of NaOH = 40 g/mole and # of moles of NaOH unreacted = 0.051

Mass of NaOH unreacted = # of moles x molar mass = 0.051 x 40 = 2.04 g

In simmilar way,

Mass of NaCl formed = 0.185 x 58.44 = 10.81 g.

Mass of H2O formed = 0.185 x 18 = 3.33 g.

Total volume of water added = 35.6 + 35.6 = 71.2 mL = 71.2 g (as density of water is 1 g/mL).

Total mass of water H2O in reaction mixture = 71.2 + 3.33 = 74.53 g.

On adding mass of all componants in reaction mixture we get,

Mass of reaction mixture = 74.53 + 2.04 + 10.81 = 87.38 g.

Mass of NaCl = 1.81 g

% Mass of NaCl = [(Mass of NaCl) / (Total mass of reaction mixture)]x 100

% Mass of NaCl = [10.81 / 87.38] x 100 = 12.37 %

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