Answer:
[tex]3.0molMg(OH)_2[/tex]
Explanation:
Hello there!
In this case, for these types of acid-base neutralizations, it is crucial to firstly set up the chemical reaction taking place between the acid and the base; in this case HCl and Mg(OH)2 respectively, whose products are obtained by switching around the anions and cations as shown below:
[tex]HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+H_2O(l)[/tex]
Which must be balanced to accurately predict the mole ratio on the reactants side:
[tex]2HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+2H_2O(l)[/tex]
Whereas we can see a 2:1 mole ratio of the acid to the base; thus, the moles of Mg(OH) required for the neutralization of 6.0 moles of HCl turn out to be:
[tex]6.0molHCl*\frac{1molMg(OH)_2}{2molHCl} \\\\=3.0molMg(OH)_2[/tex]
Best regards!
