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].
