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Murljashka [212]
2 years ago
15

Heat Transfer Lab

Chemistry
1 answer:
scoray [572]2 years ago
5 0

Heat Transfer Lab

The following represents a lab set up for heat transfer. The cup on the left started with boiling water at 100 degrees C and the cup on the right has water at 20 degrees C. There is an aluminum bar between the two cups allowing heat to transfer from one cup into the other. The set up will be left alone for 20 minutes and temperatures of each cup of water will be recorded every minute for 20 minutes.

mag-aral ka

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The questions are attached​
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Answer:

Explanation:

a) Balanced chemical equation:

C₃H₈ + 5O₂      →        3CO₂ + 4H₂O

b) Mass of oxygen gas needed = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/  44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of oxygen and propane:

                            C₃H₈        :          O₂        

                                1           :            5

                             0.57        :         5/1×0.57 =  2.85 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 2.85 mol × 32 g/mol

Mass = 91.2 g    

c) water vapors produced = ?

Mass of propane = 25 g

Solution:

Number of moles of propane = mass/molar mass

Number of moles = 25 g/  44.1 g/mol

Number of moles = 0.57 mol

now we will compare the moles of water and propane:

                            C₃H₈        :          H₂O        

                                1           :            4

                             0.57        :         4/1×0.57 =  2.28 mol

Mass of water vapors:

Mass = moles × molar mass

Mass = 2.28 mol × 18 g/mol

Mass = 41.0 g

d)

mass of water vapors produced = 1.5 kg (1500 g)

Mass of propane reacted = ?

Solution:

Number of moles of water vapors:

Number of moles = mass/molar mass

Number of moles = 1500 g/ 18 g/mol

Number of moles = 83.33 mol

now we will compare the moles of water and propane.

         

                            H₂O          :            C₃H₈    

                               4            :              1

                             83.3         :         1/4×83.3 =  20.8 mol

Mass of propane:

Mass = number of moles ×molar mass

Mass = 20.8 mol × 44.1 g/mol

Mass = 917 g

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