The chemical reaction involving Mg(OH)2 and HCl is:
Mg(OH)2 + 2HCl --> MgCl2 + 2H2O
So we see that for every 2 moles of HCl, 1 mole of Mg is reacted.
Calculating for moles HCl:
moles HCl = 0.300 M * 0.215 L
moles HCl = 0.0645 mol
The moles Mg then is:
moles Mg = 0.0645 mol * (1 / 2)
moles Mg = 0.03225 mol
0.03255 moles of Mg is used for neutralizing 0.3 M 245 ml [tex]\rm Mg(OH)_2[/tex].
The chemical reaction of the reaction of Mg with water yields Magnesium hydroxide.
The neutralization reaction of [tex]\rm Mg(OH)_2[/tex] with HCl will be:
[tex]\rm Mg(OH)_2\;+\;2\;HCl\;\rightarrow\;MgCl_2\;+2\;H_2O[/tex]
For neutralizing 1 mole of magnesium hydroxide 2 moles of HCl are required.
The moles of HCl in the question are:
moles = [tex]\rm molarity\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]
moles of HCl = 0.3 [tex]\rm \times\;\dfrac{1000}{215}[/tex]
moles of HCl = 0.0645 moles.
The moles of Mg required are half of HCl.
Moles of Mg = [tex]\rm \dfrac{0.0645}{2}[/tex]
Moles of Mg = 0.03225.
0.03255 moles of Mg is used for neutralizing 0.3 M 245 ml [tex]\rm Mg(OH)_2[/tex].
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