chemical equation of propane , C3H8+5O2 -->3CO2+4H2O How many grams of O2(g) are needed to completely burn 62.0 g of C3H8(g)?

Molar mass of C₃H₈ = (12.0×3 + 1.0×8) g/mol = 44.0 g/mol
Number of moles of C₃H₈ burnt = (62.0 g) / (44.0 g/mol) = 1.409 mol

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio C₃H₈ : O₂ = 1 : 5
No. of moles of O₂ needed = (1.409 mol) × 5 = 7.045 mol

Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
Mass of O₂ needed = (7.045 mol) × (32.0 g/mol) = 225 g (3 sig. fig.)

戇戇居士 <<< answered the problem correctly. That is probably how your teacher wants it done, or some variation of that. --- But you were taught a law .. the law of definite composition .. you can use a proportion, which I find easier ...
find molar masses of O2 and C3H8
O2 --> 32g/mole X 5moles = 160g
C3H8 --> 44g/mole X 1 mole = 44g

62 / 44 = x / 160 ... solve for x
x = 160(62/44) <<< use a calculator