1.51 g of butan-1-ol heated 300 cm3 of water by 42.0°C. Find the enthalpy change of combustion. I did q=mc(Delta)T and worked out the moles and everything but I got 19750.5 kJ mol-1 and the answer is supposed to be 2581.1 kJ mol–1.
1.51 g of butan-1-ol heated 300 cm3 of water by 42.0°C. Find the enthalpy change of combustion. I did q=mc(Delta)T and worked out the moles and everything but I got 19750.5 kJ mol-1 and the answer is supposed to be 2581.1 kJ mol–1.
E=-mc*delta T mass= 0.300 kg since volume of water is 300cm^3 c=4.18 change in temperature is 42
E= -0.3*4.18*42= - 52.668 KJ 1.51 g of butanol released 52.668 KJ 1.51 g is how many moles? moles= 1.51/74=0.0204 moles 1 mole of butanol released 52.668/0.0204= 2581.08 KJ so enthalpy change is -2581.1 KJ/mol