Answer :
Answer:
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
Answer:
85.07 mg of Mg
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