2C₃H₅(OH)₃ + 7O₂---> 6CO₂ + 8H₂OmC₃H₅(OH)₃=3*12u+8*1u+3*16u=92u 2*92g glicerolu------- 7*22,4dm3 tlenu (dane z reakcji) 92g glicerolu---------- xdm3 tlenu (dane z treści zad.) x=92*7*22,4/2*92x = 78,4dm3 tlenu
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2C₃H₅(OH)₃ + 7O₂---> 6CO₂ + 8H₂O
mC₃H₅(OH)₃=3*12u+8*1u+3*16u=92u
2*92g glicerolu------- 7*22,4dm3 tlenu (dane z reakcji)
92g glicerolu---------- xdm3 tlenu (dane z treści zad.)
x=92*7*22,4/2*92
x = 78,4dm3 tlenu