Oblicz, ile dm³ tlenu ( warunki normalne) potrzeba do całkowitego spalenia 220g propanu?
mC3H8=44u
C3H8 + 5O2 ----> 3CO2 + 4H2O
44g C3H8 ------ 5*22,4dm3 tlenu
220g C3H8 ----- xdm3 tlenu
x = 560dm3 tlenu
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mC3H8=44u
C3H8 + 5O2 ----> 3CO2 + 4H2O
44g C3H8 ------ 5*22,4dm3 tlenu
220g C3H8 ----- xdm3 tlenu
x = 560dm3 tlenu