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Answer :
To determine the products of the neutralization reaction between nitric acid and strontium hydroxide, we first recognize that a neutralization reaction between an acid and a base produces a salt and water.
1. The acid in the reaction is nitric acid ([tex]$\mathrm{HNO_3}$[/tex]).
2. The base in the reaction is strontium hydroxide ([tex]$\mathrm{Sr(OH)_2}$[/tex]).
3. The salt produced is strontium nitrate, with the formula [tex]$\mathrm{Sr(NO_3)_2}$[/tex], because the strontium ion ([tex]$\mathrm{Sr^{2+}}$[/tex]) combines with two nitrate ions ([tex]$\mathrm{NO_3^-}$[/tex]).
4. Water ([tex]$\mathrm{H_2O}$[/tex]) is the other product of the reaction.
With these products, the overall balanced chemical equation is:
[tex]$$
\mathrm{Sr(OH)_2 + 2\,HNO_3 \longrightarrow Sr(NO_3)_2 + 2\,H_2O.}
$$[/tex]
Even though the balanced equation shows coefficients to conserve mass and charge, when listing the products we express them as [tex]$\mathrm{Sr(NO_3)_2}$[/tex] (the salt) and [tex]$\mathrm{H_2O}$[/tex] (water).
Therefore, the final answer is:
[tex]$$
\mathrm{Sr(NO_3)_2 + H_2O.}
$$[/tex]
This corresponds to option 1.
1. The acid in the reaction is nitric acid ([tex]$\mathrm{HNO_3}$[/tex]).
2. The base in the reaction is strontium hydroxide ([tex]$\mathrm{Sr(OH)_2}$[/tex]).
3. The salt produced is strontium nitrate, with the formula [tex]$\mathrm{Sr(NO_3)_2}$[/tex], because the strontium ion ([tex]$\mathrm{Sr^{2+}}$[/tex]) combines with two nitrate ions ([tex]$\mathrm{NO_3^-}$[/tex]).
4. Water ([tex]$\mathrm{H_2O}$[/tex]) is the other product of the reaction.
With these products, the overall balanced chemical equation is:
[tex]$$
\mathrm{Sr(OH)_2 + 2\,HNO_3 \longrightarrow Sr(NO_3)_2 + 2\,H_2O.}
$$[/tex]
Even though the balanced equation shows coefficients to conserve mass and charge, when listing the products we express them as [tex]$\mathrm{Sr(NO_3)_2}$[/tex] (the salt) and [tex]$\mathrm{H_2O}$[/tex] (water).
Therefore, the final answer is:
[tex]$$
\mathrm{Sr(NO_3)_2 + H_2O.}
$$[/tex]
This corresponds to option 1.
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