Question

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H₂O . Suppose 28.1 g of hydrochloric acid is mixed with 17. g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

First, we must start off with a balanced equation, which conveniently, it already is
Overall reaction:

HCl + NaOH -------> NaCl + H2O

This gives us the stoichiometry of the reaction, so we can see that 1 mole of either reactant gives us 1 mole of water. The next step requires us to figure out which of them is in excess. To do so, we divide the given mass by each compound's molar mass as shown below:

Molar mass of HCl: 1.008+35.453 = 36.461 grams/mole

Molar mass of NaOH: 22.989+15.999+1.008 = 39.996 grams/mole

Molar mass of H2O = 2(1.008)+15.999 = 18.015 grams/mole

Molar mass of NaCl = 22.98+35.453 = 58.44 grams/mole

Now, to see how many moles of each we have:
Moles of HCl = 28.1/36.461 = 0.7707 moles HCl
Moles of NaOH = 17 /39.996 = 0.42504 moles NaOH

Hence, the limiting reagent in the reaction is the NaOH . Thus, the reaction cannot produce any more NaCl than what the 0.42504 moles of NaOH can react with. Now, from the stoichiometry of the reaction, we see that 0.42504 moles of NaOH will produce 0.42504 moles of NaCl . Multiplying the molar mass of NaCl by 0.42504 moles will give us the mass of the Nacl produced.

Mass of NaCl produced:
58.44 grams/mole * 0.4250425 moles = 24.83948 grams of NaCl

Now onto the significant figures...

Since the calculations we used relied upon the NaOH , the answer should come out to 3 significant figures.

So, the maximum mass of NaCl produced is equal to:
Mass of NaCl = 24.84 grams