\(M_{CO_2}=12+16.2=44\left(\dfrac{g}{mol}\right)\)
\(n_{CO_2}=\dfrac{m}{M}=\dfrac{44}{44}=1\left(mol\right)\)
\(V_{CO_2\left(đktc\right)}=n.22,4=1.22,4=22,4\left(l\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{44}{44}=1\left(mol\right)\\n_{Cl_2}=\dfrac{7,1}{71}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{hh}=\left(1+0,1\right).22,4=24,64\left(lít\right)\)