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

Explanation :

First we have to calculate the moles of and NaI.

and,

The balanced chemical reaction will be,

From the balanced reaction we conclude that,

As, 1 mole of neutralizes by 2 mole of NaI

So, 0.3 mole of 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 =

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

where,

q = heat absorbed = ?

= specific heat of water =

m = mass of water = 450 g

= final temperature of water =

= initial temperature of metal =

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

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