combustion reaction...practice exam question?

Question

Consider the combustion reaction:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
When 0.200 mol of C3H8 undergoes combustion, the heat released is sufficient to raise the temperature of 1550 g of water from 20.0 °C to 70.8 °C. Based on these data, calculate ΔH for this reaction. Specific heat of water is 4.18 J/ g•°C. All choices are per mol of C3H8.

A. -1650 molkJ
B. -1240 molkJ
C. -860 molkJ
D. -1950 molkJ
E. -430 molkJ

How did you work it out? Thanks.

Answer 1

Find the heat liberated by the reaction:

Q = m*c*ΔT, Q = 1550*4.18*(70.8 - 20), Q = 329,133.2 J

Now if the combustion of 0.2 mol of C3O8 liberates 329.133.2 J the combustion of 1 mol gives: 329,133.2/0.2 = 1,645,666 J or 1,646 kJ of heat. So.

ΔH = -1646 kJ/mol or -1650 kJ/mol approx.

Answer 2

Recall Q = m Cp Del T

Therefore

Heat = 1550 x 4.18 x (70.8 - 200 / 0.2 =1645 kJ

It will have negative sign as there is a release of energy and to 3 sig figures -1650 kJ/mol

Answer A

I think you have your units wrong in your answer selection.