Question

The balanced combustion reaction for C 6 H 6 is 2 C 6 H 6 ( l ) 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) 6 H 2 O ( l ) 6542 kJ If 7.700 g C 6 H 6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water

Answer 1

Answer:
`156.4^0C`

Step-by-step explanation:

`2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542kJ`
`\Delta H=-6542kJ`

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=(7.700g)/(78.11g/mol)=0.9858moles`

According to stoichiometry :

2 moles of
`C_6H_6` releases = 6542 kJ of heat

0.9858 moles of
`C_6H_6` release =
`(6542)/(2)* 0.9858=3224kJ` of heat

Thus heat given off by burning 7.700 g of
`C_6H_6` will be absorbed by water.

`Q=m* c* \Delta T`

Q = Heat absorbed= 3224kJ = 3224000J (1kJ=1000J)

m= mass of water = 5691 g

c = specific heat capacity =
`4.184J/g^0C`

Initial temperature of the water =
`T_i` = 21.0°C

Final temperature of the water =
`T_f` = ?

Putting in the values, we get:

`3224000J =5691g* 4.184J/g^0C* (T_f-21)`

`T_f=156.4^0C`

The final temperature of the water is
`156.4^0C`