Consider the following precipitation reaction:

To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -Molar mass H2SO4 = 98g/mol- we can find the mass of sulfuric acid produced:

Moles sulfuric acid:

8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4

Mass sulfuric acid:

4mol H2SO4 * (98g / mol) =

392g sulfuric acid are produced

4 0

3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.

Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.

The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.

From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

So, the answer is:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

5 0

3 years ago