\(m_{O_2}=15.04-11.2=3.85\left(g\right)\)
\(n_{O_2}=\dfrac{3.85}{32}=0.1203125\left(mol\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
Bảo toàn e :
\(n_{SO_2}=\dfrac{3\cdot0.2-0.1203125\cdot4}{2}=0.059375\left(mol\right)\)
\(V_{SO_2}=1.33\left(l\right)\)