Question

The combustion of how many moles of ethane (C2H6) would be required to heat 851 g of water from 25.0°C to 98.0°C? (Assume liquid water is formed during the combustion.)

Answer 1

Answer : The number of moles of ethane required will be 0.166 mole.

Explanation :

First we have to calculate the heat absorbed by water.

`q=m* c* (T_(final)-T_(initial))`

where,

q = heat absorbed = ?

m = mass of water = 851 g

c = specific heat of water =
`4.18J/g^oC`

`T_(final)` = final temperature =
`98.0^oC`

`T_(initial)` = initial temperature =
`25.0^oC`

Now put all the given values in the above formula, we get:

`q=851g* 4.18J/g^oC* (98.0-25.0)^oC`

`q=259674.14J=259.67kJ` (1 kJ = 1000 J)

Now we have to calculate the moles of ethane required.

`\Delta H=(q)/(n)`

where,

`\Delta H` = enthalpy of combustion of ethane = 1560.7 kJ/mol (standard value)

q = heat absorbed = 259.67 kJ

n = number of moles of ethane = ?

`1560.7kJ/mol=(259.67kJ)/(n)`

`n=(259.67kJ)/(1560.7kJ/mol)`

`n=0.166mole`

Therefore, the number of moles of ethane required will be 0.166 mole.