Answer : The number of moles of ethane required will be 0.166 mole.
Explanation :
First we have to calculate the heat absorbed by water.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat absorbed = ?
m = mass of water = 851 g
c = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]98.0^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]25.0^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=851g\times 4.18J/g^oC\times (98.0-25.0)^oC[/tex]
[tex]q=259674.14J=259.67kJ[/tex] (1 kJ = 1000 J)
Now we have to calculate the moles of ethane required.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy of combustion of ethane = 1560.7 kJ/mol (standard value)
q = heat absorbed = 259.67 kJ
n = number of moles of ethane = ?
[tex]1560.7kJ/mol=\frac{259.67kJ}{n}[/tex]
[tex]n=\frac{259.67kJ}{1560.7kJ/mol}[/tex]
[tex]n=0.166mole[/tex]
Therefore, the number of moles of ethane required will be 0.166 mole.
