a) \(H=H_1.H_2.H_3=100\%.75\%.80\%=60\%\)
b) \(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(kmol\right)\)
BTNT S: \(n_{FeS_2}=\dfrac{1}{2}n_{H_2SO_4}=0,25\left(kmol\right)\)
=> \(m=\dfrac{0,25.120}{60\%}=50\left(kg\right)\)
a)
Hiệu suất toàn quá trình : $H = 100\%.75\%.80\% = 60\%$
b)
$n_{H_2SO_4} = \dfrac{49}{98}= 0,5(kmol)$
$4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
$2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$SO_3 + H_2O \to H_2SO_4$
Theo PTHH, $n_{FeS_2\ pư} = \dfrac{1}{2}n_{H_2SO_4} = 0,25(kmol)$
$\Rightarrow n_{FeS_2\ đã\ dùng} = 0,25 : 60\% = \dfrac{5}{12}(kmol)$
$m_{FeS_2} = \dfrac{5}{12}.120 = 50(kg)$
$\%FeS_{2_{trong\ quặng}} = 100\% - 10\% = 90\%$
$\Rightarrow m_{quặng} = 50 : 90\% = 55,56(kg)$
