Răspuns:
3molii.......8moli......9moliiii
3Fe 304+8Ai->(t>2000°C) 4Ai203+9Fe
0.1moli......,0.88moli.......x moli...........
Nr moli=m/M=23.2/232=0.1 moli Fe 304
MFe 304=56×3+16×4=232 g/mol
Nr moli=m/M=23.76/27=0.88 moli
MAi=27 g/mol
M=4.95×56=277.2g Fe