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\(n_{Al}=\frac{108\cdot1000}{27}=4000\left(mol\right)\)
PTHH : \(2Al_2O_3\underrightarrow{dpnc\left(criolit\right)}4Al+3O_2\)
Theo PTHH : \(n_{Al_2O_3}=\frac{1}{2}n_{Al}=2000\left(mol\right)\)
=> \(m_{Al_2O_3}=2000.102=204000\left(g\right)=204\left(kg\right)\)
Mà hiệu suất phản ứng là 80% => \(m_{Al_2O_3\left(thực\right)}=\frac{204}{80}\cdot100=255\left(kg\right)\)
=> \(m_{quặng}=\frac{255}{50}\cdot100=510\left(kg\right)\)
\(n_{Fe} = \dfrac{5000.1000}{56} = \dfrac{625000}{7}\ kmol\\ n_{FeS_2\ đã\ dùng} = \dfrac{n_{Fe}}{H\%} = \dfrac{\dfrac{625000}{7}}{89,6\%} = 99649,23\ kmol\\ m_{quăng\ pirit} = \dfrac{m_{FeS_2}}{90\%} = \dfrac{99649,23.120}{90\%} = 13259897,33 (kg) = 13259,89(tấn)\)
\(n_{Fe}=\dfrac{5000\cdot10^6}{56}=\dfrac{625}{7}\cdot10^6\left(mol\right)\)
\(BTFe:\)
\(n_{FeS_2}=n_{Fe}=\dfrac{625}{7}\cdot10^6\left(mol\right)\)
\(n_{FeS_2\left(tt\right)}=\dfrac{\dfrac{625}{7}\cdot10^6}{89.6}=\dfrac{56000\cdot10^6}{7}\left(mol\right)\)
\(\Rightarrow m_{FeS_2}=\dfrac{56000\cdot10^6\cdot120}{7}=960000\cdot10^6\left(g\right)=960000\left(tấn\right)\)
\(m_{quặng}=\dfrac{960000\cdot100}{90}=1066666.67\left(tấn\right)\)
a)
$4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
$2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$SO_3 + H_2O \to H_2SO_4$
b)
$m_{FeS_2} = 1000.60\% = 600(kg)$
$n_{FeS_2} = \dfrac{600}{120} = 5(kmol)$
$n_{FeS_2\ pư} = 5.80\% = 4(kmol)$
$n_{H_2SO_4} = 2n_{FeS_2} = 8(kmol)$
$m_{H_2SO_4} = 8.98 = 784(kg)$
4FeS2+11O2to→2Fe2O3+8SO2
4FeS2+11O2→to2Fe2O3+8SO2
2SO2+O2to,xt−−→2SO3
2SO2+O2→to,xt2SO3
SO3+H2O→H2SO4
SO3+H2O→H2SO4
b)
mFeS2=1000.60%=600(kg)
mFeS2=1000.60%=600(kg)
nFeS2=600120=5(kmol)
nFeS2=600120=5(kmol)
nFeS2 pư=5.80%=4(kmol)
nFeS2 pư=5.80%=4(kmol)
nH2SO4=2nFeS2=8(kmol)
nH2SO4=2nFeS2=8(kmol)
mH2SO4=8.98=784(kg)
4FeS2 + 11O2 \(\underrightarrow{to}\) 2Fe2O3 + 8SO2 (1)
2SO2 + O2 \(\underrightarrow{to}\) 2SO3 (2)
SO3 + H2O → H2SO4 (3)
\(m_{FeS_2}=80\times40\%=32\left(kg\right)=32000\left(g\right)\)
\(n_{FeS_2}=\frac{32000}{120}=\frac{800}{3}\left(mol\right)\)
Theo PT1: \(n_{SO_2}=2n_{FeS_2}=2\times\frac{800}{3}=\frac{1600}{3}\left(mol\right)\)
Theo PT2: \(n_{SO_3}=n_{SO_2}=\frac{1600}{3}\left(mol\right)\)
Theo PT3: \(n_{H_2SO_4}=n_{SO_3}=\frac{1600}{3}\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=\frac{1600}{3}\times98=52266,67\left(g\right)=52,26667\left(kg\right)\)
4FeS2 + 11O2 \(\underrightarrow{t^o}\) 2Fe2O3 + 8SO2 (a)
2SO2 + O2 \(\underrightarrow{t^o}\) 2SO3 (b)
SO3 + H2O \(\underrightarrow{t^o}\) H2SO4 (c)
\(m_{FeS_2}\) = 80 . 40% = 32 ( kg ) = 32000 ( g)
\(n_{FeS_2}\) = \(\dfrac{32000}{120}\) = \(\dfrac{800}{3}\) (mol )
Theo phương trình (a) có : \(n_{SO_2} = 2n_{FeS_2}= 2. \dfrac{800}{3} = \dfrac{1600}{3} ( mol )\)
Theo phương trình (b) có : \(n_{SO_3}= n_{SO_2} = \dfrac{1600}{3} (mol)\)
Theo phương trình (c) có : \(n_{H_2SO_4} = n_{SO_3} = \dfrac{1600}{3} (mol)\)
\(\rightarrow\) \(m_{H_2SO_4} = \dfrac{1600}{3} . 98 = 52266 , 67 (g) = 52 , 26667 (kg)\)