Question

An electrical heater is used to add 21.75 kJ of heat to aconstant-volume calorimeter. The temperature of the calorimeterincreases by 4.18 oC. When 3.00 g of butanol(C4H9OH) is burned in the same calorimeter,the temperature increases by 20.81 oC. Calculate themolar heat of combustion for butanol.

Answer 1

Answer:

The molar heat of combustion for butanol is 2.7 *10³ kJ/mol

Explanation:

Step 1: Data given

Temperature change = 4.18 °C

mass of butanol = 3.00 grams

When butanol is burned Temperature change = 20.81 °C

Step 2: Calculate Ccalorimeter

Ccal = 21750J / 4.18 °C = 5203.35 J/ °C

Step 3: Calculate heat generated

Heat generated = 20.81 °C * 5203.35 J/°C = 108281.71 J = 108.28 kJ

Step 4: Calculate moles of butanol

Number of moles = mass / molar mass

Number of moles = 3 grams / 74.123 g/mol = 0.04 moles

Step 5: Calculate Molar heat of combustion

Molar heat of combustion = 108.28 kJ / 0.04 moles = 2707 kJ /mol

The molar heat of combustion for butanol is 2.7 *10³ kJ/mol