How many moles of [tex]$Mg(OH)_2$[/tex] are needed to completely neutralize [tex]$2.48 \, \text{mol} \, HNO_3$[/tex]?

[tex]\boxed{\text{mol} \, Mg(OH)_2}[/tex]



Answer :

To determine how many moles of [tex]\( \mathrm{Mg(OH)_2} \)[/tex] are needed to completely neutralize [tex]\( 2.48 \)[/tex] moles of [tex]\( \mathrm{HNO_3} \)[/tex], we can follow these steps:

1. Write down the balanced chemical equation for the reaction between magnesium hydroxide ([tex]\( \mathrm{Mg(OH)_2} \)[/tex]) and nitric acid ([tex]\( \mathrm{HNO_3} \)[/tex]):
[tex]\[ \mathrm{Mg(OH)_2 + 2HNO_3 \rightarrow Mg(NO_3)_2 + 2H_2O} \][/tex]

2. From the balanced equation, observe the molar ratio between [tex]\( \mathrm{Mg(OH)_2} \)[/tex] and [tex]\( \mathrm{HNO_3} \)[/tex]. It shows that:
[tex]\[ 1 \text{ mole of } \mathrm{Mg(OH)_2} \text{ neutralizes } 2 \text{ moles of } \mathrm{HNO_3} \][/tex]

3. Given that we have [tex]\( 2.48 \)[/tex] moles of [tex]\( \mathrm{HNO_3} \)[/tex], and knowing the stoichiometric relationship, we can set up a proportion to find the moles of [tex]\( \mathrm{Mg(OH)_2} \)[/tex] needed:

[tex]\[ 2.48 \text{ moles of } \mathrm{HNO_3} \times \frac{1 \text{ mole of } \mathrm{Mg(OH)_2}}{2 \text{ moles of } \mathrm{HNO_3}} = \text{moles of } \mathrm{Mg(OH)_2} \][/tex]

4. Calculate the moles of [tex]\( \mathrm{Mg(OH)_2} \)[/tex] needed:

[tex]\[ \frac{2.48}{2} = 1.24 \text{ moles of } \mathrm{Mg(OH)_2} \][/tex]

Thus, [tex]\( 1.24 \)[/tex] moles of [tex]\( \mathrm{Mg(OH)_2} \)[/tex] are needed to completely neutralize [tex]\( 2.48 \)[/tex] moles of [tex]\( \mathrm{HNO_3} \)[/tex].