Percent yield of chemical reactions

Question

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O

If 0.370g of water is produced from the reaction of 1.1g of hydrochloric acid and 2.0g of sodium hydroxide, calculate the percent yield of water.

Prompt assistance is appreciated.

Answer 1

Hello, Kristen,

We need the balanced equation for this reaction, which is:

1HCl + 1NaOH = 1NaCl + 1H₂O

This tells us we should expect 1 mole of H₂O for every 1 mole of starting reactants, which are also in a 1 to 1 molar ratio.

No calculate the moles of HCl and BaOH thjat we are given by dividing their masses by their molar masses:

O

HCl: 1.2gram/(36.45 grams/mole) = 0.0320 moles HCl

NaOH: 2.0g/(40.0g/mole) = 0.0500 moles NaOH.

Note that we have more NaOH than we do of the HCl. We need equimolar amounts, according to the balanced equation. Once the 0.0302 moles of HCl is consumed by the same number of moles of NaOH, the HCl will be gone and we'll have 0.0198 moles of NaOH left over.

The 0.0302 moles of HCl should produce 0.0302 moles of H₂O. Convert that into grams by multiplying by water's molar mass of 19.0 grams/mole. That gives us 0.543 grams of water. But we only got 0.370 grams H₂O. That is a yield of (0.37/0.543) or 0.681. That's a 68.1% yield.

Bob