Hydrogen chloride obtained by the action of sulfuric acid on NaCl weighing 58.5 g was dissolved in water weighing 146 g

Hydrogen chloride obtained by the action of sulfuric acid on NaCl weighing 58.5 g was dissolved in water weighing 146 g. Calculate the mass fraction (%) of HCl in the solution.

When sodium chloride and concentrated sulfuric acid react with slight heating, hydrogen chloride gas and sodium sulfate are formed. The reaction occurs according to the equation:

2 NaCl + H2SO4 = 2 HCl + Na2SO4

Let’s calculate the mass of the released hydrogen chloride.

Molar mass of hydrogen chloride = 36 g / mol, molar mass of sodium chloride = 58 g / mol. Let’s make the proportion:

58.5 g of NaCl corresponds to X g of HCl, as

2 * 58 g / mol NaCl correspond to 2 * 36 g / mol HCl.

X = (58.5 * 2 * 36) / (2 * 58.5) = 36.31 g of hydrogen chloride

Now we will find the mass fraction of hydrogen chloride in an aqueous solution. To do this, first of all, we calculate the mass of the solution:

m (solution) = 36.31 + 146 = 180.31

Now let’s find the mass fraction:

w = (36.31 * 100) / 180.31 = 20%

Answer: mass fraction 20%.