Câu 1 :
\(2KMnO_4+16HCl_{\left(đ\right)}\underrightarrow{^{t^0}}2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(H_2+Cl_2\underrightarrow{^{as}}2HCl\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
Câu 2 :
\(n_{Mg}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)
\(m_{hh}=24a+65b=14.55\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=a+b=0.35\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.2,b=0.15\)
\(\%m_{Mg}=\dfrac{0.2\cdot24}{14.55}\cdot100\%=33.98\%\)
\(\%m_{Zn}=100-33.98=67.01\%\)
Câu 3 :
\(m_{FeS_2}=187.5\cdot80\%=150\left(kg\right)\)
\(n_{FeS_2}=\dfrac{150\cdot10^3}{120}=1250\left(mol\right)\)
\(n_{H_2SO_4}=2n_{FeS_2}=1250\cdot2=2500\left(mol\right)\)
\(n_{H_2SO_4\left(tt\right)}=2500\cdot90\%=2250\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{2250\cdot98\cdot100}{98}=225000\left(g\right)\)
\(V_{dd_{H_2SO_4}}=\dfrac{225000}{1.84}=122282\left(ml\right)=122.282\left(l\right)\)
