# Question 32fe8

Dec 11, 2017

We're given an assumption and data to arrive at the conclusion. First we need to calculate the heat released from cooling water,

$q = m {C}_{s} \Delta T$

q = 17.2g * (4.18J)/(g*°C) * (-80°C) * (kJ)/(10^3J) approx -5.75kJ

of heat are released by cooling that water.

The equation for combusting butane is a combustion reaction,

$2 {C}_{4} {H}_{10} + 13 {O}_{2} \to 8 C {O}_{2} + 10 {H}_{2} O$ where DeltaH^° approx -5.75kJ#

This is where many will make a stupid mistake, there are 2 moles of butane in that reaction, so our calculations must take that into account when calculating the grams needed. I guarantee you there would be an answer on the test for this mistake.

$\frac{- 5.75 k J}{2 \text{mol") approx (-2.88kJ)/("mol}}$

Moving on, we need,

$- 350 k J \cdot \left(\text{mol")/(-2.88kJ) * (58g)/("mol}\right) \cdot \frac{k g}{{10}^{3} g} \approx 7.05 k g$

of butane to release the quantity of heat desired.