Question

Question #c8d5a

Answer 1

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

`(8.17kJ)/("mol")`

Hence,

`45.7kJ = n * (8.17kJ)/("mol")`
`therefore n approx 5.60"mol" * (16.0g)/("mol") = 89.5g`

of methane may be boiled using that amount of heat.