Answer:
744.2 g of C2H6 must burn to raise the temperature of 39.0 L of water by 58.0°C
Explanation:
This excersise is about calorimetry.
Q = m . C . ΔT
For water, 58°C is the ΔT, and the specific heat is 4.18 kJ/kg°C. We are missing the mass, but with the density data, we can know it.
Water density = water mass / water volume
1 g/ml = water mass / 39000 mL
(Notice we had to convert 39 L in mL (.1000))
Water mass = 39000 g
But this is in grams, and we have to make again a conversion, to kg because the units of specific heat.
Q = 39 kg . 4.18 kJ/ kg.°C . 58°C
Q = 9455.16 kJ
This is the heat required to change water temperature with that water mass, and the heat released for one mol of C2H6 is 382kJ.
How many mol of C2H6, for the heat required to change water, need the chemist?. The rule of three will be:
382 kJ ____ 1 mol of C2H6
9455.16 kJ _____ (9455.16 / 382) = 24.7 moles of C2H6
For mass, just work with the molar weight.
Mol . molar weight = mass
24.7 mol . 30.07g/m =744.2 g
