Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 31.4 g of hydrochloric acid is mixed with 12. g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

The answer to your question is there will be 20.5 g left of HCl

Explanation:

Data

HCl = 31.4 g

NaOH = 12 g

Excess of HCl = ?

Balanced chemical reaction

                HCl  +  NaOH  ⇒   NaCl  +  H₂O

Molar weight of HCl = 1 + 35.5 = 36.5 g

Molar weight of NaOH = 23 + 16+ 1 = 40 g

Calculate the limiting reactant

theoretical yield HCl/NaOH = 36.5/40 = 0.9125

experimental yield HCl/NaOH = 31.4/12 = 2.62

Conclusion

The limiting reactant is NaOH because the experimental yield increases.

Calculate the mass of Excess reactant

               36.5 g of HCl 40 g of NaOH

                x                     12 g of NaOH

                x = (12 x 36.5) / 40

                x = 438 / 40

                x = 10.95 g of HCl

Excess HCl = 31.4 - 10.95

                   = 20.5 g

There will remain 20.5 grams of hydrochloric acid

Explanation:

Step 1: Data given

Mass of hydrochloric acid = 31.4 grams

Mass of sodium hydroxide = 12.0 grams

Molar mass hydrochloric acid (HCl) = 36.46 g/mol

Molar mass sodium hydroxide (NaOH) = 40.0 g/mol

Step 2: The balanced equation

HCl + NaOH → NaCl + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles HCl = 31.4 grams / 36.46 g/mol

Moles HCl = 0.861 moles

Moles NaOH = 12.0 grams / 40.0 g/mol

Moles NaOH = 0.3 moles

Step 4: Calculate the limiting reactant

The limiting reactant is NaOH. It will completely be consumed (0.3 moles). Hcl is in excess. There will react 0.3 moles. There will remain 0.861 - 0.3 = 0.561 moles

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass

Mass HCl = 0.561 moles * 36.46 g/mol

Mass HCl = 20.5 grams

There will remain 20.5 grams of hydrochloric acid

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