Question

How many moles of Mg(OH)2(aq) would be required to neutralize 6.0 mol HCl(aq)?

Answer 1

Answer:

$3.0molMg(OH)_2$

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:

$HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+H_2O(l)$

Which must be balanced to accurately predict the mole ratio on the reactants side:

$2HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+2H_2O(l)$

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:

$6.0molHCl*\frac{1molMg(OH)_2}{2molHCl} \\\\=3.0molMg(OH)_2$

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