Oblicz masę tlenu potrzebnego do spalenia 104 g. etynu do sadzy i pary wodnej .
2C2H2+O2---->4C+2H2O
104g x
2mC2H2=4*12u+4*1u=52u
2mO=2*16u=32u
52u--------104g
32u--------x
x=64g tlenu
2C₂H₂ + O₂ -----> 4C + 2H₂O
mO=16u
mO₂=2*16u=32u
mC₂H₂=2*12u+2*1u=24u+2u=26u
m2C₂H₂=2*26u=52u
32g tlenu - 52g etynu
xg tlenu - 104g etynu
x=(32g tlenu*104g etynu)/52g etynu=64g tlenu
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2C2H2+O2---->4C+2H2O
104g x
2mC2H2=4*12u+4*1u=52u
2mO=2*16u=32u
52u--------104g
32u--------x
x=64g tlenu
2C₂H₂ + O₂ -----> 4C + 2H₂O
mO=16u
mO₂=2*16u=32u
mC₂H₂=2*12u+2*1u=24u+2u=26u
m2C₂H₂=2*26u=52u
32g tlenu - 52g etynu
xg tlenu - 104g etynu
x=(32g tlenu*104g etynu)/52g etynu=64g tlenu