M,MnO2= 87g/mol
M,Mn=55g/mol
M,Al= 27g/mol----> niu=m./M= 135/27mol=5mol
3mol..............4mol............2mol............3mol
3MnO2 + 4Al ------> 2Al2O3 + 3Mn
x.....................5.mol......................................y
x= 15/4molMnO2-------> m=niuxM= 15/4X87g=326,25 g oxid pur
p=90%= m,px100/m,i---> m,i= 326,25x100/90=g=362,5 g oxid impur
y= 15/4 mol Mn----> m= niuxM= 15/4x55g=206,25g Mn ,este masa teoretica, cand randamentul este 100%;cum exista pierderi, o sa se obtina mai putin
85/100= m,practic/206,25------> calculeaza m,pr !!!!!!!