Question

the following equations is the balanced combustion reactions for c6h6

2C6H6(l)+15O2---->12CO2(G)+6H20(l)+6542 k j

if 7.700 g of C6H6is 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)= 7.700 g

use:

number of mol of C6H6,

n = mass of C6H6/molar mass of C6H6

=(7.7 g)/(78.11 g/mol)

= 9.858*10^-2 mol

Since Δ H is negative, heat is released

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

So,

for 9.858*10^-2 mol of C6H6, heat released = 9.858*10^-2*6542.0/2 KJ

= 322.5 KJ

= 322500 J

This heat is absorbed by water

Given:

Q = 322500 J

m = 5691 g

C = 4.184 J/g.oC

Ti = 21 oC

use:

Q = m*C*(Tf-Ti)

322500 = 5691*4.184*(Tf-21)

Tf -21.0 = 13.5 oC

Tf = 34.5 oC

Answer: 34.5 oC