ta có :
n_{cl2} = 5.\frac{13.5}{5.71+2.160}=0.1 (mol); n_{NaI} = \frac{36}{150}=0.24
Phản ứng :
cl_2 + 2NaI ------> 2Nacl + I_2
0.1---->0.2\\\\\\\\\\\ 0.2
Br_2 + 2NaI ------->2NaBr + I_2
0.02<--0.04--------->0.04
m = m_{Nacl} + m_{NaBr} = 0,2 .58,5 +103.0,04=15,82 g