The questions are attached

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

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.