Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°c into steam at 101.0°c. (assume that the h2o produced in the combustion reaction is steam rather than liquid water.)

The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

Now, we calculate the amount of heat released or absorbed in all the processes.

where,

= amount of heat absorbed = ?

m = mass of water = 242.0 g

= specific heat of water = 4.18 J/g°C

= final temperature =

= initial temperature =

Putting all the values in above equation, we get:

where,

= amount of heat absorbed = ?

m = mass of water or steam = 242 g

= latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

where,

= amount of heat absorbed = ?

m = mass of steam = 242.0 g

= specific heat of steam = 2.08 J/g°C

= final temperature =

= initial temperature =

Putting all the values in above equation, we get:

Total heat required =

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be =

To calculate the number of moles, we use the equation:

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

Hence, the mass of methane burned is 12.4 grams.

i belive it is water