oblicz objętość tlenu w warunkach normalnych , niezbędną do całkowitego spalania 60 g propynu
mC3H4=3*12u+4*1u=40u
C3H4 + 4 O2---->3 CO2 + 2 H2O
40g C3H4-------3*22,4dm3 tlenu
60g C3H4---------xdm3 tlenu
x = 100,8dm3 tlenu
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mC3H4=3*12u+4*1u=40u
C3H4 + 4 O2---->3 CO2 + 2 H2O
40g C3H4-------3*22,4dm3 tlenu
60g C3H4---------xdm3 tlenu
x = 100,8dm3 tlenu