Respuesta :
According to the reaction equation shown, if the amount of the reactants is halved, the amount of H2O(l) produced in the reaction will be halved.
The molecular reaction equation was given as follows; HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l). Ionically, we can write; H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) ----> Na^+(aq) + + Cl^-(aq) + H2O(l). The net ionic equation now becomes; H^+(aq) + OH^-(aq) ----> H2O(l).
This implies that the concentration of H2O(l) depends on the concentrations of H^+(aq) and OH^-(aq) from HCl(aq) and NaOH(aq) respectively and the mole ratio of reactants and products is 1:1:1:1.
Hence, of the amount of the reactants is halved, the amount of H2O(l) produced in the reaction will be halved.
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Titration is the quantitative analytical method in which the concentration of a substance is known by the measured titrant. According to the question, the amount of H[tex]_2[/tex]O produced will be halved, when the reactants will be halved.
The molecular equation of sodium hydroxide, hydrochloric acid, sodium chloride, and water can be represented as:
- [tex]\text{HCl + NaOH} \rightarrow \text{NaCl + H}_2\text O[/tex]
In the given equation, spectator ions and ionic equation can be identified as:
- [tex]\text{H}^+ +\text{Cl}^- +\text{Na}^+ +\text{OH}^- \rightarrow\;\text{Na}^+ +\text{Cl}^-+\text{H}_2 \text{O}[/tex]
The net ionic equation will be:
- [tex]\text{H}^++ \text{OH}^- \rightarrow\text{H}_2 \text{O}[/tex]
Now, from the above equation it can be observed that the production of water molecule is dependent on the number of reactants. The reactants are
- H[tex]\textH^+[/tex] and [tex]\text{OH}^-[/tex].
Therefore, if the number of reactants are halved, then the resulting water molecule will be halved.
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