miu,KCl03= (39gK+35,5gCl+3x16gO)/mol= 122,5g/mol
niu,mol= m/miu; niu= 1225g/122,5g/mol= 10mol KClO3
ecuatia reactiei
1mol............1mol.....1,5mol
KClO3----> KCl + 3/2O2
10mol.........................niu=15mol
deci, teoretic s-ar obtine 15mol iar practic s-a obtinut mai putin, 12mol
randament= 12x100/15=> calculeaza !!!!