Question

Hydrogen chloride can be made in the labatory by the reaction of sodium chloride and sulfric acid!

___NaCl(s)+______H2SO4(aq)____HCl(aq)_____Na2SO4(aq)

Blance the above equation

If 35. g of NaCl are used , How many grams of H2SO4 are required for a complete reaction

If 35. g of NaCl are used , and only 8.6g of HCl are isolated what is the percentage yield of HCl

If only 8.6 g of HCl are obtained , How many grams of Na2SO4, must have been formed.

Answer 1

Reaction:

NaCl (s) + H2SO4 (aq) --- > HCl (aq) + Na2SO4 (aq)

Lets balance the given reaction

2 NaCl (s) + H2SO4 (aq) --- > 2 HCl (aq) + Na2SO4 (aq)

1)

Mole ration of NaCl to H2SO4

Calculation of moles of NaCl = 35.0 g x 1 mol / 58.44 g

= 0.599 mol NaCl

Moles of H2SO4 required = 0.599 mol NaCl x 1 mol H2SO4 / 2 mol NaCl

= 0.2994 mol H2SO4

Mass of H2SO4 = 0.2994 mol H2SO4 x molar mass of H2SO4

= 0.2994 mol H2SO4 x 98.079 g/mol

= 29.36 g

2).

Lets calculate theoretical yield of HCl

Moles of HCl formed from 0.599 mol NaCl

= 0.599 mol NaCl x 2 mol HCl / 2 mol NaCl

= 0.599 mol HCl

Mass of HCl = 0.599 mol HCl x 36.46 g/mol

= 21.84 g

Percent yield = actual yield / theoretical yield x 100

= 8.6 g / 21.84 g * 100

= 39.4 %

3).

Mole of HCl = 8.6 g / 36.46 g per mol = 0.236 mol HCl

Moles of Na2SO4 = 0.236 mol HCl x 1 mol Na2SO4 / 2 mol HCl

= 0.118 mol

Mass of Na2SO4 = 0.118 mol x molar mas of Na2SO4

= 0.118 mol x 142.04 g per mol

= 16.75

Mass of sodium sulfate formed = 16.75 g