Enthalpy problem 2?

Question

Burning butane(C2H10) produces gaseuous carbon dioxide and water. The enthalpy of combustion of butane is -2650 kJ per mole. Determine how much water you can heat from room temperature (22 degrees C) to boiilng with 1 lb of butane.

WILL RATE YOU 10 POINTS

Answer 1

6 liters de agua

Answer 2

Butane, BTW, is C4H10.

So, you have 1 lb (453.59 g) of C4H10.  To figure out how much water you're going to heat from 22 deg C to  boiling, you need to:

1. Figure out how much heat the butane's going to produce

2. Figure out how much water that's going to boil (Q = mc DT)

Step 1:

Combustion of butane produces -2650 kJ/mol.  How many moles of C4H10 are in 453.59g?

1 mol C4H10 = (12.01g *4) + (1.01g *10) = 58.14 g

453.59g * (1 mol/ 58.14g ) = 7.81 mol

-2650 kJ/mol (7.81 mol) = -20674 kJ

Which means that -20764 kJ of heat is released in the combustion of 1 lb butane.

Step 2:

All of the heat (presumably) from combusting 1 lb butane is used to heat up the water, so the amount of heat that is required is +20764 kJ.  (same number, but different sign to denote heat absorbed)

So:

Q = mC(delta T)

Q/ (C* (delta T) ) = m

Q = 20764 kJ

C = 4.18 kJ/ (kg*C) [note: you may have learned this with units of J/g*C.  The number itself is still the same if you changed J and g to kJ and kg.  Trust me.]

delta T = 100C - 22C = 78C

20764 kJ/ (4.18 [kJ/ (kg*C)] * 78C) = m

63 kg = m