oblicz objętość tlenu niezbędną do spalania całkowitego 60-ciu gr. propynu.
mC3H4=3*12u+4*1u=40u
C3H4 + 4 O2--->3CO2 + 2H2O
40g C3H4---------4*22,4dm3 tlenu
60g C3H4-----------xdm3 tlenu
x = 134,4dm3 tlenu
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mC3H4=3*12u+4*1u=40u
C3H4 + 4 O2--->3CO2 + 2H2O
40g C3H4---------4*22,4dm3 tlenu
60g C3H4-----------xdm3 tlenu
x = 134,4dm3 tlenu