Ta có:
\(m_C=1.90\%=0,9=900g\)
\(\Rightarrow n_C=\dfrac{900}{12}=75\left(mol\right)\)
Theo gt ta có PTHH: \(C+O_2-t^o->CO_2\) (*)
Theo (*) và gt có: 75mol...75mol......75mol
\(\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=75.44=3300\left(g\right)\\m_{O_2}=75.32=2400\left(g\right)\\V_{O_2}=75.22,4=1680\left(l\right)\end{matrix}\right.\)
Vì \(V_{kk}=5.V_{O_2}\) nên \(V_{kk}=5.1680=8400\left(l\right)\)