Respuesta :
Answer:
57.1 kilo Joules of heat would be released.
Explanation:
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex] ΔH°=-57.1kJ/mol
Molarity of HCl = 2.00 M
Molarity of NaOH = 1.00 M
According to reaction , 1 M of HCl reacts with 1 M of NaOH. Then 2.00 M of HCl will react with:
[tex]\frac{1}{1}\times 2.00 M= 2M [/tex] of NaOH
But according to question we only have 1.00 M NaOH .So, this means that NaOH is limiting reagent and HCl is an excessive reagent.
Heat evolved will depend upon concentration of NaOH solution :
Heat evolved when 1.00 M of NaOH reacts =
[tex]1.00\times (-57.1 kJ/mol)=-57.1 kJ[/tex]
Negative sign means that heat is released during the reaction.
57.1 kilo Joules of heat would be released.
The heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.
The given reaction,
[tex]\bold {HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l) \ \ \ \ \ \ \ \ \ \Delta H = - 57.1\ kJ/mol}[/tex]
Concentration of HCl is 2 mole.
So, 2 mole of HCl react with 2 mole of NaOH but only mole of NaOH is available.
So, NaOH is a Limiting factor in the reaction and HCl is excessive factor.
Only, 1mole of HCl will react with available 1 mole of NaOH,
Therefore, the heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.
To know more about Enthalpy,
https://brainly.com/question/9444545
