Question

150.0 mL of 2.00 M Pb(NO3)2 solution is mixed with 300.0 mL of 2.00 M NaI solution in a coffee cup calorimeter of negligible heat capacity.
The initial temperature of the two solutions are both at 20.00 oC, the final temperature of the mixed solution is 36.09 oC.

c of H 2O = 4.184 J/g oC ; d of solution is 1.00g/mL. Refer to the following:
Pb(NO3)2(aq) + 2NaI (aq) --> PbI2(s) + 2NaNO3(aq)
What is q of the water? Please watch units. Answers have units of kJ. You have to convert J to kJ.

-1790 kJ/mol

42.9 kJ/mol

- 451 kJ/mol

30.3 kJ/mol

20.9 kJ/mol

91.7 kJ/mol

85.4 kJ/mol

Answer 1

Answer : The value of q of the water is, 30.3 kJ/mol

Explanation :

First we have to calculate the moles of $Pb(NO_3)_2$ and NaI.

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}=2.00mole/L\times 0.150L=0.3mole$

and,

$\text{Moles of }NaI=\text{Concentration of }NaI\times \text{Volume of solution}=2.00mole/L\times 0.300L=0.6mole$

The balanced chemical reaction will be,

$Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3$

From the balanced reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ neutralizes by 2 mole of NaI

So, 0.3 mole of $Pb(NO_3)_2$ 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 = $150.0mL+300.0mL=450mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 450ml=450g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.184J/g^oC$

m = mass of water = 450 g

$T_{final}$ = final temperature of water = $20.0^oC$

$T_{initial}$ = initial temperature of metal = $36.09^oC$

Now put all the given values in the above formula, we get:

$q=450g\times 4.184J/g^oC\times (36.09-20.0)K$

$q=30294.252J=30.3kJ$

Therefore, the value of q of the water is, 30.3 kJ/mol