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The questions are attached​

The questions are attached​-example-1

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Answer:

Step-by-step 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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