# Question c8d5a

Dec 31, 2017

We'll need a bit of data to solve this. The enthalpy of vaporization of methane is approximately,

$\frac{8.17 k J}{\text{mol}}$

Hence,

$45.7 k J = n \cdot \frac{8.17 k J}{\text{mol}}$
therefore n approx 5.60"mol" * (16.0g)/("mol") = 89.5g#

of methane may be boiled using that amount of heat.