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DaniilM [7]
3 years ago
14

To begin the experiment, 1.5g of methane CH4 is burned in a bomb calorimeter containing 1,000 grams of water. The initial temper

ature of water is 25°C. The specific heat of water is 4.184 J/g°C. The heat capacity of the calorimeter is 695 J/°C . After the reaction the final temperature of the water is 37°C.
1. Using the formula q =m• C• ΔT ,calculate the heat absorbed by the water. Show work.

2. The total heat absorbed by the water and the calorimeter can be calculated by adding the heat calculated in steps 3 and 4. The amount of heat released by the reaction is equal to the amount of heat absorbed with the negative sign as this is an exothermic reaction. Using the formula ∆H = -(qcal + qwater), calculate the total heat of combustion. Show work.

3. Calculate the experimental molar heat of combustion in kJ/mol? Show work.

4. Given the formula: % error = ((accepted value-experimental value)/accepted value) * 100, calculate the percent error for the experiment. Show work
Chemistry
1 answer:
vredina [299]3 years ago
4 0

Mass of methane takne = 1.5g

moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles


mass of water = 1000 g

Initial temperature of water = 25 C

final temperature = 37 C

specific heat of water = 4.184 J /g C

1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) =  50208 Joules

2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J

3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules

This heat is released by 0.094 moles of methane

So heat released by one mole of methane =

                            - 622851.06 Joules = 622.85 kJ / mole

4) standard enthalpy of combustion = -882 kJ / mole

Error = (882-622.85) X 100 / 882 = 24.84 %

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2.5 moles of KOOH are produced.

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1)Given data:

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now we will compare the moles of KOOH and LiClO.

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