Respuesta :
Answer : The value of q of the water is, 30.3 kJ/mol
Explanation :
First we have to calculate the moles of [tex]Pb(NO_3)_2[/tex] and NaI.
[tex]\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}=2.00mole/L\times 0.150L=0.3mole[/tex]
and,
[tex]\text{Moles of }NaI=\text{Concentration of }NaI\times \text{Volume of solution}=2.00mole/L\times 0.300L=0.6mole[/tex]
The balanced chemical reaction will be,
[tex]Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3[/tex]
From the balanced reaction we conclude that,
As, 1 mole of [tex]Pb(NO_3)_2[/tex] neutralizes by 2 mole of NaI
So, 0.3 mole of [tex]Pb(NO_3)_2[/tex] neutralizes by 0.6 mole of NaI
Thus, the number of neutralized moles = 0.6 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water = [tex]150.0mL+300.0mL=450mL[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 450ml=450g[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat absorbed = ?
[tex]c[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]
m = mass of water = 450 g
[tex]T_{final}[/tex] = final temperature of water = [tex]20.0^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature of metal = [tex]36.09^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=450g\times 4.184J/g^oC\times (36.09-20.0)K[/tex]
[tex]q=30294.252J=30.3kJ[/tex]
Therefore, the value of q of the water is, 30.3 kJ/mol
