m(Fe*C)=90g=100% "2Fe+6HCl=2FeCl3+3H2" - reactia ,, s uglerodom toest s "C" v reactie ne vstupaet n(H2)=V/V(molar)=33,6l/22,4l/mol=1.5mol ,, n(Fe)=1,5mol*2mol/3mol=1mol
m(Fe)=M*n=1mol*56g.mol= 56g,,, m(C)=90g-56g=34g,,,,,w(Fe)=56g*100%/90g=62,(2)%~62%,,,,,,,,,,,,,w(C)=34g*100%/90g=37,(7)%~38% nu ili mojno bilo opredelit po drugomu doliu carbona tak kak Fe*Cl eto 100% to w(C)=100%-62%=38% poluceatsea tojesamoe vibirai kakoi hocesh))..................... R:w(Fe)=62%,w(C)=38%