oblicz objętość tlenu niezbędną do całkowitego spalania 60 g propynu z góry dzięki :)
C3H4 =12g*3+4*1g =40g Reakcja spalania propynu: 60g X(dm3) C3H4+4O2→3CO2+2H2O 40g 4*22,4dm3 60g X 40g 4*22,4dm3 60g*4*22,4dm3 X= = 134,4dm3 40g
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C3H4 =12g*3+4*1g =40g
Reakcja spalania propynu:
60g X(dm3)
C3H4+4O2→3CO2+2H2O
40g 4*22,4dm3
60g X
40g 4*22,4dm3
60g*4*22,4dm3
X= = 134,4dm3
40g