The combustion of what volume of ethane (C2H6), measured at 23.0°C and 752 mmHg, would be required to heat 855 g of water from 25.02°C to 98.02°C?

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

a c d and f

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