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 of hydrochloric acid and of sodium hydroxide? Round your answer to significant figures.

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 33.2 g of hydrochloric acid and 13.1 g of sodium hydroxide? Round your answer to 3 significant figures.

Answer:

x ≈ 19.6 g (Three significant Figures)

Explanation:

First thing's first, we have to write out the chemical equation.

HCl (aq) + NaOH (s) --------> NaCl (aq) + H2O (l)

Next up, we have to determine the limiting reactant. We do this by comparing the number of moles of the reactants.

Number of moles = Mass / Molar mass

The molar masses are given below as;

HCl = 36.458 g/mol

NaOH = 39.997 g/mol

Number of moles of HCl = 33. 2 / 36.458 = 0.911 mol HCl

Number of moles of NaOH = 13.1 / 39.997 = 0.328 mol NaOH

So its clear that the limiting reactant is NaOH. We have an excess of HCl.

1 mol of NaOH produces 1 mol of NaCl

39.997g (1mol * 39.997g/mol) of NaOH produces 58.44g (1mol * 58.44 g/mol) of NaCl

Note: Mass = Number of moles * Molar mass

13.1 g of NaoH would produce xg?

38.997= 58.44

13.1 = x

x = (13.1 * 58.44 ) / 38.997 = 19.63136g

x ≈ 19.6 g (Three significant Figures)