Question

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

Answer 1

85.07 mg of Mg

Step-by-step explanation:

no of moles of HCl

= 1.032*0.219

= 0.226 mol

after the reaction No of moles of HCl left

= 1*0.219 = 0.219 mol

No of moles of HCl reacted = 0.226-0.219

= 0.007 mol

Mg(s)+2HCl→MgCl2(aq)+H2(g)

from the reaction 1mole Mg = 2mole HCl

No of moloes of Mg = 0.007/2

= 0.0035 mol

mass of Mg = 0.0035*24.3

= 0.0851 grams

= 85.07 mg of Mg

Answer 2

85.07 mg of Mg must be added

Step-by-step 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