a/ \(n_{SO_2\left(đktc\right)}=\dfrac{V}{22,4}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)
\(4FeS_2+11O_2\rightarrow^{t^0}2Fe_2O_3+8SO_2\uparrow\)
4 11 2 8 (mol)
0,8 2,2 0,4 1,6 (mol)
\(m_{FeS_2}=0,8.120=96\left(g\right)\)
b/ \(m_{tc}=100-m_{FeS_2}=100-96=4\left(g\right)\)
\(\Rightarrow\%m_{tc}=\dfrac{4.100}{100}\%=4\%\)
c/ \(m_{Fe_2O_3}=0,4.160=64\left(g\right)\)
\(\Rightarrow m=m_{tc}+m_{Fe_2O_3}=4+64=68\left(g\right)\)
d/ \(V_{O_2\left(lt\right)}=2,2.22,4=49,28\left(l\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=56-V_{O_2\left(lt\right)}=56-49,28=6,72\left(l\right)\)
\(\Rightarrow V_{hh}=V_{O_2\left(dư\right)}+V_{SO_2}=6,72+35,84=42,56\left(l\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{O_2\left(dư\right)}=\dfrac{6,72}{42,56}.100\%\approx15,79\%\\\%V_{SO_2}=\dfrac{35,84}{42,56}.100\%\approx84,21\%\end{matrix}\right.\)
