Trả lời chủ đề

Gọi n[SUB]MOH[/SUB]=n[SUB]M2CO3[/SUB]=1(mol) -> nMCl=3(mol), n[SUB]CO2[/SUB]=1 (mol)

-> n[SUB]H+[/SUB]=3(mol) -> m[SUB]ddHCl[/SUB]= (3.36,5.100)/14,6=750 (g)

-> m[SUB]dd sau[/SUB]= 750+M+17+2M+60-44  = 3M+783

-> C% MCl = 3(M+35,5)/(3M+783)=0,43656 <-> M=139 ( có vẻ vô lý thật)


Top