Can you please help me I don't understand this.?

1)
Molar mass of C₃H₈ = (12.0×3 + 1.0×8) g/mol = 44.0 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol

Initial moles of C₃H₈ = (3.25 g) / (44.0 g/mol) = 0.07386 mol
Initial moles of O₂ = (3.50 g) / (32.0 g/mol) = 0.1094 mol

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio C₃H₈ : O₂ = 1 : 5

When O₂ completely reacts, C₃H₈ needed = (0.1094 mol) × (1/5) = 0.02188 mol < 0.07386 mol
Hence, C₃H₈ is in excess.
The limiting reactant is O₂.

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2)
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol

According to the equation in 1), mole ratio O₂ : H₂O = 5 : 4
Moles of H₂O formed = ((0.1094 mol) × (4/5) = 0.08752 mol
Mass of H₂O formed = (0.08752 mol) × (18.0 g/mol) = 1.58 g

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3.
Refer to 1).
Initial moles of C₃H₈ = 0.07386 mol
Moles of C₃H₈ reacted = 0.02188 mol
Moles of excess C₃H₈ remains after reaction = (0.07386 - 0.02188) mol = 0.05198 mol

Mass of excess C₃H₈ remains after reaction = (0.05198 mol) × (44.0 g/mol) = 2.29 g

Propane + Oxygen ➜ Carbon Dioxide + water
C₃H₈ + 5O₂ ➜ 3CO₂ + 4H₂O
molecular weights
C = 12
H = 1
O = 16
C₃H₈ = 3•12+8 = 44
5O₂ = 10•16 = 160
3CO₂ = 3(12+32) = 132
4H₂O = 4•18 = 72
check 44 + 160 = 132 + 72 = 204

1 mole of C₃H₈ + 5 moles of O₂ ➜ 3 mole of CO₂ + 4 moles of H₂O
44 grams of C₃H₈ + 160 grams of O₂ ➜ 132 grams of CO₂ + 72 grams of H₂O

3.25 g of C3H8 react with 3.50 g of O2
Write the formula for the reactant that is the limiting reactant
ratio of C₃H₈ to O₂ is 44/160
you have 3.25/3.5
not enough O₂, you would need 
3.25•160/44 = 11.8 g
so O₂ is the limiting reactant

How many grams of H2O are formed?
ratio of H₂O to O₂ is 72/160
72/160 = x/3.50
x = 1.575 g

How much of the excess reactant, in grams, remains after the reaction?
grams of C₃H₈ used:
ratio is O₂ to C₃H₈ is 44/160
44/160 = x/3.5 
x = 0.9625
difference is 3.25 – 0.9625 = 2.29 g