Answer :
Answer:
4.11L
Explanation:
To heat the mass of water from 25.02°C to 98.02°C we need:
Q = C*m*ΔT
Where Q is heat, C is specific heat of water (4.184J/mol°C), m is mass of water (855g) and ΔT is change in temperature (98.02°C - 25.02°C = 73°C)
Q = C*m*ΔT
Q = 4.184J/mol°C*855g*73°C
Q = 261144J = 261.14kJ
In the combustion of 1 mole of ethane there are released 1560kJ/mol.
To release 261.14kJ there are necessaries:
261.14kJ * (1mol / 1560kJ) = 0.1674 moles
Using PV = nRT
We can find the volume that 0.1674 moles of ethane at 23.0°C and 752mmHg occupy.
Temperature: 273.15 + 23°C = 296.15K
Pressure: 752mmHg * (1atm / 760mmHg) = 0.9895atm
V = nRT / P
V = 0.1674mol*0.082atmL/molK * 296.15K / 0.9895atm
V = 4.11L