Oblicz, ile dm^3 CO powstanie w reakcji spalania 45g propanu w tlenie w warunkach normalnych.
2C3H8 + 7O2 -----> 6CO + 8H2O
88g C3H8 --------- 134,4dm3 CO
45g C3H8 --------- X
X= 68,73dm3 CO
mC3H8=44u
2C3H8 + 7 O2---->6CO + 8H2O88g C3H8----------6*22,4dm3 CO45g C3H8------------xdm3 COx = 68,73dm3 CO
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2C3H8 + 7O2 -----> 6CO + 8H2O
88g C3H8 --------- 134,4dm3 CO
45g C3H8 --------- X
X= 68,73dm3 CO
mC3H8=44u
2C3H8 + 7 O2---->6CO + 8H2O
88g C3H8----------6*22,4dm3 CO
45g C3H8------------xdm3 CO
x = 68,73dm3 CO