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* \text{Volume of solution}=2.00mole/L* 0.150L=0.3mole`

and,

`\text{Moles of }NaI=\text{Concentration of }NaI* \text{Volume of solution}=2.00mole/L* 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}* \text{Volume of water}=1g/ml* 450ml=450g`

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

`q=m* c* (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* 4.184J/g^oC* (36.09-20.0)K`

`q=30294.252J=30.3kJ`

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