ta có : \(m_C=\frac{96\cdot1}{100}=0,96\left(kg\right)=960\left(g\right)\Rightarrow n_C=\frac{960}{12}=80\left(mol\right)\\ m_S=\frac{4\cdot1}{100}=0,04\left(kg\right)=40\left(g\right)\Rightarrow\\ n_S=\frac{40}{32}=1,25\left(mol\right)\\ \)
Ta có PTHH:
\(C+O_2\rightarrow CO_2\\ \Rightarrow n_{O_2}=n_{CO_2}=n_C=80\left(mol\right)\\ \Rightarrow m_{CO_2}=80\cdot44=3520\left(g\right)\\ \\ S+O_2\rightarrow SO_2\\ \Rightarrow n_{O_2}=n_S=1,25\left(mol\right)\\ \\ \)
Ta có \(V_{O_2}=\left(80+1,25\right)\cdot22,4=81,25\cdot22,4=1820\left(l\right)\)