Respuesta :
Answer : The enthalpy change per mole of zinc reacting for the reaction is 152.4 kJ/mol.
Explanation :
First we have to calculate the moles of Zn and HCl.
[tex]\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}[/tex]
Molar mass of Zn = 65 g/mole
[tex]\text{Moles of }Zn=\frac{1.34g}{65g/mole}=0.0206mole[/tex]
and,
[tex]\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.750M\times 0.0600=0.0450mole[/tex]
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Zn[/tex] react with 2 mole of [tex]HCl[/tex]
So, 0.0206 moles of [tex]Zn[/tex] react with [tex]0.0206\times 2=0.0412[/tex] moles of [tex]HCl[/tex]
From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Zn[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the enthalpy change per mole of zinc reacting for the reaction.
From the reaction we conclude that,
As, 0.0206 moles of Zn produces heat = 3.14 kJ
So, 1 mole of Zn produces heat = [tex]\frac{3.14kJ}{0.0206mol}=152.4kJ/mol[/tex]
Therefore, the enthalpy change per mole of zinc reacting for the reaction is 152.4 kJ/mol.
The enthalpy change per mole of zinc reacting for the reaction is 152.43 KJ/mol
We'll begin by calculating the number of mole of in 1.34 g Zn.
Mass of Zn = 1.34 g
Molar mass of Zn = 65 g/mol
Mole of Zn =?
Mole = mass / molar mass
Mole of Zn = 1.34 / 65
Mole of Zn = 0.0206 mole
Next, we shall determine the number of mole of HCl in 60.0 mL of 0.750 M HCl
Molarity of HCl = 0.750 M
Volume of solution = 60 mL = 60 / 1000 = 0.06 L
Mole of HCl =?
Mole = Molarity × Volume
Mole of HCl = 0.75 × 0.06
Mole of HCl = 0.045 mole
Next, we shall determine the limiting reactant.
Zn(s) + 2HCl(aq) —> ZnCl₂(aq) + H₂(g)
From the balanced equation above,
1 mole of Zn reacted with 2 moles of HCl.
Therefore,
0.0206 mole of Zn will react with = 0.0206 × 2 = 0.0412 mole of HCl
From the above calculation, we can see that only 0.0412 mole of HCl out of 0.045 mole that was given reacted completely with 0.0206 mole of Zn.
Thus, Zn is the limiting reactant and HCl is the excess reactant.
Finally, we shall the enthalpy change per mole of zinc reacting for the reaction.
From the reaction,
0.0206 mole of Zn reacted to produce 3.14 kJ of heat.
Therefore, 1 mole of Zn will react to produce = [tex]\frac{3.14}{0.0206} \\\\[/tex] = 152.43 KJ of heat energy.
Thus, the enthalpy change per mole of zinc reacting for the reaction is 152.43 KJ/mol
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