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m glicerolu = 100g
V O = ?
C3H5(OH)3 + 2O2 → 3CO + 4H2O
92u + 64u → 84u + 72u
92g C3H5(OH)3 - 64 g O2
100g C3H5(OH)3 - 69,6 g O2
m O = 69,6 g
V = 69,6 ÷ 1,43 = 48,7 dm³
Odp. Potrzebna objętość tlenu to 48,7 dm³.