2Me + 3Cl2 ⇒ 2MeCl3
n MeCl3 = n Me
1,522/(A+3·35,5) = 0,754/A
1,522/(A+106,5) = 0,754/A
1,522A = 0,754A + 80,301
0,768A = 80,301 A = 104,55g/mol ...... nu prea exista in sistemul periodic un asemenea metal ...⇒ text scris gresit , dar, ......mod de rezolvare corect