Oblicz objętość tlenu w warunkach normalnych niezbędną do całkowitego spalenia 60g propynu.
C3H4 + 4O2------> 3CO2 + 2H2O
60g------xdm{3}
40g------4*22,4dm3
x=.....dm3
40g(masa C3H4)
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C3H4 + 4O2------> 3CO2 + 2H2O
60g------xdm{3}
40g------4*22,4dm3
x=.....dm3
40g(masa C3H4)