Question

H₂SO₄ + NaCl → Na₂SO₄ + HCl, Given 5.09 g of H₂SO₄ and 6.33 g of NaCl

1.Calculate the mass (in g) of the excess reagent remaining after the reaction is complete.

2.Calculate the theoretical yield of HCl (in g) from the reaction

3.Determine the percent yield of HCl if 1.91 g of HCl is actually produced from the reaction.

Answer 1

The balanced reaction is:

$H_2SO_4+2\, NaCl \rightarrow Na_2SO_4+2\, HCl$

Molar mass of H₂SO₄ is 98 g.mol^-1 , thus moles of it was used (5.09 / 98) mol = 0.0519 mol / /

Molar mass of NaCl is 58.5 g.mol^-1 , thus moles of it was used (6.33 / 58.5) mol = 0.1082 mol

1. According to the mole ratio of them (H₂SO₄ : NaCl = 1:2), moles of NaCl should be twice of that of H₂SO₄. Thus, moles of NaCl should be (2 x 0.0519) mol = 0.1038 mol. So, NaCl is in excess and after complete reaction it will remain (0.1082 - 0.1038) mol = 0.0044 mol = (0.0044 x 58.5) g = 0.2574 g.

2. H₂SO₄ is the limiting reagent here. So, yield has to be measured based on amount of it. Since moles of HCl produced is twice the moles of H₂SO₄. Thus, theoretical moles of HCl should be (2 x 0.0519) mol = 0.1038 mol = (0.1038 x 36.5) g = 3.7887 g.

3. Percentage yield = (amount produced / amount should be produced) x % = (1.91 / 3.7887) x 100% = 50.4 %.