2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
MC₂H₆ = 2*12 + 6*1 = 24 + 6 = 30 (g)
MO₂ = 2*16 = 32 (g)
2*30 g C₂H₆ -------- 7*32 g O₂
12 g C₂H₆ -------- x g O₂
x = [tex]\frac{7*32*12}{2*30}[/tex] = [tex]\frac{2688}{60}[/tex] = 44,8 (g O₂)
Odp: Z 12 gramami etanu przereaguje 44,8 gramów tlenu.
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
MC₅H₁₂ = 5*12 + 12*1 = 60 + 12 = 72 (g)
MO₂ = 2*16 = 32 (g)
72 g C₅H₁₂ -------- 8*32 g O₂
12 g C₅H₁₂ -------- x g O₂
x = [tex]\frac{8*32*12}{72}[/tex] = [tex]\frac{3072}{72}[/tex] ≈ 42,7 (g O₂)
Odp: Z 12 gramami pentanu przereaguje 42,7 gramów tlenu.
