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.

$q=m\times c\times (T_{final}-T_{initial})$

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:

$q=851g\times 4.18J/g^oC\times (98.0-25.0)^oC$

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

Now we have to calculate the moles of ethane required.

$\Delta H=\frac{q}{n}$

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 = ?

$1560.7kJ/mol=\frac{259.67kJ}{n}$

$n=\frac{259.67kJ}{1560.7kJ/mol}$

$n=0.166mole$

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