68.5g of C3H8 are reacted with 83.4g of O2 in the following reaction. how many grams of the excess reactant must remain unreacted

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

mole C₃H₈ = 68.5 : Mr C₃H₈ = 68.5 : 44 g/mol = 1.557

mole O₂ = 83.4 : Mr O₂ = 83.4 : 32 g/mol = 2.606

to find limiting reactants :

divide the number of moles by their respective coefficients, and the small quotient will be a limiting reactants

C₃H₈ : O₂ = 1.557/1 : 2.606/5 = 1.557 : 0.521

O₂ as a limiting reactant, C₃H₈ excess reactants

reacted mole of C₃H₈ : 1/5 x 2.606 = 0.5212

unreacted mole of C₃H₈ : 1.557 - 0.5212 = 1.0358

mass of C₃H₈ (unreacted) = 1.0358 x 44/gmol = 45.575 g

i think its 100.

hope this !