100% ......................... 800g carbid impur
80% ......................... mp = 640 g carbid pur
640 g n moli
CaC2 + 2H2O → Ca(OH)2 + C2H2(g)
64 1
=> n = 640x1/64 = 10 moli C2H2 sau obtinut la 100%... dar la 70%... cat..?
100% .......................... 10 moli
70% ........................... y = 7 moli C2H2
nr.moli = V/Vm, Vm = 22,4
=> V = nxVm = 7x22,4 = 156,8 L C2H2