DATA GIVEN

Mass of C₅H₁₀ = 35.00g
Mass of O₂ = 45.00g
Asked:
Mass of excess reagent remain after the reaction.
Chemical equation
C₅H₁₀ + 17/2O₂ → 6CO₂ + 5H₂O

Limitting Reagent

To convert masses into moles
Mole = mass /molar mass
Mole of C₅H₁₀ = 35.00g /70g/mole
= 0.5 mole

Mole ratio for C₅H₁₀ = Mole reacted /mole from equation (mole coefficient)
= 0.5 mole /1mole
= 0.5 mole

Mole for O₂ = Mass /molar mass
= 45.00g /32g/mole
= 1.40625 mole
Mole ratio for O₂ = mole reacted /mole coefficient
Mole ratio = 1.40625 /17/2
= 1.40625 × 2 /17
= 0.1654411

The limiting reagent is Oxygen and excess reagent is hexgne (C₆H₁₀)

Mass of excess reagent

Mass of excess remain is
1 mole of C₆H₁₀ Ξ 17/2 mole of O₂
? Ξ 1. 406

= 0.165 mole of C₆H₁₀
Mass = mole × molar mass
= 0.165mole × 70g/mole
= 11.5788g
Mass remain = Original mass – reacted mass
= 35g – 11.5788g
= 23.42g
Mass of excess reagent remain unreacted was 23.42g