Question

The following equation is the balanced combustion reaction for C6H6:

2C6H6(l)+15O2(g) = 12CO2(g)+6H2O(l)+6542KJ

If 7.800 g of C6H6 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

Molar mass of C6H6,

MM = 6*MM(C) + 6*MM(H)

= 6*12.01 + 6*1.008

= 78.108 g/mol

mass(C6H6)= 7.800 g

use:

number of mol of C6H6,

n = mass of C6H6/molar mass of C6H6

=(7.8 g)/(78.11 g/mol)

= 9.986*10^-2 mol

Since Δ H is negative, heat is released

when 2 mol of C6H6 reacts, heat released = 6542.0 KJ

So,

for 9.986*10^-2 mol of C6H6, heat released = 9.986*10^-2*6542.0/2 KJ

= 3.266*10^2 KJ

= 3.266*10^5 J

This heat is absorbed by water

Q = m*C*(Tf-Ti)

3.266*10^5 J = 5691 g * 4.184 J/g.oC * (Tf - 21) oC

Tf - 21 = 13.7

Tf = 34.7 oC

Answer: 34.7 oC