Oblicz, ile gramów węglika wapnia potrzeba do otrzymania 11,2dm sześciennego acetylenu w warunkach normalnych.
M(Cac2)=40+2*12=64g/mol
CaC2+2H2O--->C2H2+Ca(OH)2
64g=22,4dm^3
x=11,2dm^3
x=64^11,2/22,4 = 32g
CaC₂ + 2H₂O -----> C₂H₂ + Ca(OH)₂
MCa=40g/mol
MC=12g/mol
MCaC₂=40g/mol+2*12g/mol=40g/mol+24g/mol=64g/mol
64g CaC₂ - 22,4dm³ C₂H₂
xg CaC₂ - 11,2dm³ C₂H₂
x=(64g CaC₂*11,2dm³ C₂H₂):22,4dm³ C₂H₂=32g CaC₂
Odp. Potrzeba 32g węglika wapnia.
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M(Cac2)=40+2*12=64g/mol
CaC2+2H2O--->C2H2+Ca(OH)2
64g=22,4dm^3
x=11,2dm^3
x=64^11,2/22,4 = 32g
CaC₂ + 2H₂O -----> C₂H₂ + Ca(OH)₂
MCa=40g/mol
MC=12g/mol
MCaC₂=40g/mol+2*12g/mol=40g/mol+24g/mol=64g/mol
64g CaC₂ - 22,4dm³ C₂H₂
xg CaC₂ - 11,2dm³ C₂H₂
x=(64g CaC₂*11,2dm³ C₂H₂):22,4dm³ C₂H₂=32g CaC₂
Odp. Potrzeba 32g węglika wapnia.