Oblicz ile kg 80% karbidu potrzeba do otrzymania 1m(3) acetylenu..???

CaC2 + 2H2O → Ca(OH)2 + C2H2 M C2H2 = 2*12+2*2=24+4=28g/mol MCaC2 = 40+2*12=40+24=64 g/mol V C2H2 =1m³=1000 dm³ n=v/V0 = 1000/22,4 = 44,64 mola C2H2 1 mol CaC2 ------------------1 mol C2H2 x ----------------------------44,64 mola C2H2 x= 44,64 mola CaC2 m=n*M = 44,64 mola*64g/mol = 2857,14g CaC2 100% ------------2857,14g 80% ----------------x x=2285,71 g 80% CaC2