A method of lowering the concentration of HCl(aq) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to 219 mL of 1.032 M HCl to reduce the solution concentration to exactly 1.000 MHCl

85.07 mg of Mg must be added

Explanation:

The reaction of HCl(aq) with Mg is:

2HCl(aq) + Mg → MgCl₂ + H₂

219 mL of 1.032 M HCl are:

0.219L × (1.032mol / L) = 0.226 moles of HCl

As you want to reduce the concentration to 1.000M HCl and volume of solution is 0.219L, moles of HCl you want are 0.219 moles

That means you need to make react:

0.226 moles - 0.219 moles = 7x10⁻³ moles of HCl

Based on the reaction, 2 moles of HCl react with 1 mol of Mg, that means you need:

7x10⁻³ moles of HCl ₓ (1 mol Mg / 2 mol HCl) = 3.5x10⁻³ moles of Mg. In miligrams:

3.5x10⁻³ moles of Mg ₓ (24.305g / 1mol) ₓ (1000 mg / 1g) =

85.07 mg of Mg must be added

i think it's c. i'm not sure but i think so.