Question

If 88.31 mL of sulfuric acid was added to a mixture of 25.66g of zinc and sodium is reacted with a stoichiometric amount of sulfuric acid. The reaction mixture is then reacted with 159 mL of 3.08 M barium chloride to produce the maximum possible amount of barium sulfate. What was the concentration of the sulfuric acid

Answer 1

To find the concentration of sulfuric acid, we need to work backwards from the stoichiometry of the reaction. We know that zinc and sodium react with sulfuric acid in a 1:2 ratio, forming zinc sulfate and sodium sulfate, respectively.

So, for every 1 mol of zinc we have 2 mol of sodium. The molar mass of zinc is 65.38 g/mol and the molar mass of sodium is 23.00 g/mol.

We also know that the mass of sodium is 25.66 g, and the amount of sulfuric acid added is 88.31 mL.

The reaction between zinc and sulfuric acid has a stoichiometric coefficient of 1, so for every mol of zinc we have 1 mol of zinc sulfate.

The reaction between sodium and sulfuric acid also has a stoichiometric coefficient of 1, so for every 1 mol of sodium we have 1 mol of sodium sulfate.

So now we have a system of equations:

65.38g zinc ÷65.38g/mol =1 mol zinc

=2mol zinc sulfate × 65.38g/mol zinc sulfate =130.76g zinc sulfate

23.00g sodium ÷23.00g/mol =1 mol sodium

=2mol sodium sulfate × 77.37g/mol sodium sulfate =154.74g sodium sulfate

88.31mL sulfuric acid ÷98.08g/mL sulfuric acid =0.901 mol sulfuric acid

Using the information we have, we can set up the following equations:

y + z + n = 0.901 mol

y + 2n = 154.74 g

2z + n = 130.76 g

where y is the amount of zinc, z is the amount of sodium, and n is the amount of sulfuric acid.

Solving this system of equations, we find that y = 56.66g, z = 66.31g, and n = 60.00g.