Đặt n_{NaBr}=x(mol);n_{NaI}=y(mol)
Br_2+2NaI->2NaCl+I_2
Theo PT: n_{Br_2}=n_{I_2}=0,5y(mol)
->m_{giam}=0,5.254y-0,5.160y=7,05
->y=0,15(mol)
Cl_2+2NaBr->2NaCl+Br_2
Cl_2+2NaI->2NaCl+I_2
Theo PT:
n_{Br_2}=0,5x(mol);n_{I_2}=0,5y=0,075(mol)
n_{Cl_2}=0,5(x+y)=0,5x+0,075(mol)
->m_{giam}=160.0,5x+0,075.254-71(0,5x+0,075)=22,625
->x=0,2(mol)
->\%m_{NaI}={0,15.150}/{0,15.150+0,2.103}.100\%\approx 52,2\%
->\%m_{NaBr}=100-52,2=47,8\%