2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O If 2.56 g of C4H10 reacts with 3.45g of oxygen which is the limiting reagent?

Butane + Oxygen ➜ Carbon Dioxide + water

2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O

molecular weights

C = 12

H = 1

O = 16

2C₄H₁₀ = 2•58 = 116

13O₂ = 13•32 = 416

8CO₂ = 8•44 = 352

10H₂O = 10•18 = 180

check 116+416 = 352+180 = 532

116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O

2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O

If 2.56 g of C₄H₁₀ reacts with 3.45g of oxygen which is the limiting reagent?

amount of oxygen needed to react with 2.56 g of butane is

416/116 = x/2.56

x = 9.18 g of oxygen needed

you have only 3.45 g so it is the limiting factor

(2.56 g C4H10) / (58.1222 g of C4H10/mol) = 0.044045 mol C4H10

(3.45 g O2) / (31.99886 g O2/mol) = 0.107816 mol O2

0.107816 mole of O2 would react completely with 0.107816 x (2/13) = 0.016587 mole of C4H10, but there is more C4H10 present than that, so C4H10 is in excess and O2 is the limiting reagent.