\(Cl_2\left(0,1\right)+2NaI\left(0,2\right)\rightarrow2NaCl\left(0,2\right)+I_2\left(0,1\right)\)
\(n_{Cl_2}=\frac{2,24}{22,4}=0,1\)
\(m_{NaI}=100.45\%=45\)
\(\Rightarrow n_{NaI}=\frac{45}{150}=0,3\)
Vì \(\frac{n_{NaI}}{2}=\frac{0,3}{2}>0,1=n_{Cl_2}\)
Nên Cl2 phản ứng hết NaI dư
\(n_{NaI\left(dư\right)}=0,3-0,2=0,1\)
\(\Rightarrow m_{NaI\left(dư\right)}=0,1.150=15\)
\(m_{NaCl}=0,2.58,5=11,7\)
\(m_{Cl_2}=0,1.71=7,1\)
\(m_{I_2}=0,1.254=25,4\)
\(\Rightarrow m_{dd\left(spu\right)}=100+7,1-25,4=81,7\)
\(\Rightarrow C\%\left(NaI\right)=\frac{15}{81,7}.100\%=18,36\%\)
\(\Rightarrow C\%\left(NaCl\right)=\frac{11,7}{81,7}.100\%=14,32\%\)
