a)
\(m_{FeS_2\left(bđ\right)}=1.5\%=0,05\left(tấn\right)=50000\left(g\right)\)
=> \(n_{FeS_2\left(pư\right)}=\dfrac{50000.80\%}{120}=\dfrac{1000}{3}\left(mol\right)\)
PTHH: \(4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\)
Theo PTHH: \(n_{Fe_2O_3}=\dfrac{500}{3}\left(mol\right)\Rightarrow m_{Fe_2O_3}=\dfrac{500}{3}.160=\dfrac{80000}{3}\left(g\right)\)
b)
Theo PTHH: \(n_{SO_2}=\dfrac{2000}{3}\left(mol\right)\Rightarrow V_{SO_2}=\dfrac{2000}{3}.22,4=\dfrac{44800}{3}\left(l\right)\)