Gọi \(\left\{{}\begin{matrix}n_{O2}:a\left(mol\right)\\n_{O3}:b\left(mol\right)\end{matrix}\right.\)
\(PTHH:O_3+2KI+H_2O\rightarrow O_2+I_2+2KOH\)
\(\overline{M_A}=10,24.4=40,96\)
Sau khi dẫn qua KI , \(\Rightarrow V_{O2}=\frac{50}{22,4}\left(mol\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}a+b=\frac{50}{22,4}\\\frac{32a+48b}{a+b}=40,96\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{55}{56}\\b=1,25\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{O2}=\frac{55}{56}.22,4=22\left(l\right)\\V_{O3}=1,25.22,4=28\left(l\right)\end{matrix}\right.\)
Gọi a là số mol của O3 thêm
\(\overline{M}=10,667.4=42,668\)
\(\Rightarrow\frac{\frac{55}{56}.32+48\left(1,25+a\right)}{\frac{50}{22,4}+a}=42,668\)
\(\Leftrightarrow48\left(1,25+a\right)+\frac{220}{7}=42,668\left(\frac{50}{22,4}+a\right)\)
\(\Leftrightarrow a=0,715\)
\(\Rightarrow V_{O3}=0,715.22,6=16,0,16\left(l\right)\)
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