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nKMnO4 = 14,2/158 ≃ 0,0899 mol
2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0,0899 \(\dfrac{0,0899\times5}{2}\)
→ nCl2 = 0,22475 mol → VCl2 = 22,4.nCl2 = 5,0344 lít
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ PTHH:Fe+S\underrightarrow{t^o}FeS\left(1\right)\\ LTL:0,2>0,1\Leftrightarrow Fe.dư\)
\(Theo.pt\left(1\right):n_{Fe\left(pư\right)}=n_{FeS}=0,1\left(mol\right)\\ n_{FeS\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
\(PTHH:FeS+H_2SO_4\rightarrow FeSO_4+H_2S\uparrow\left(2\right)\\ Fe+2H_2SO_4\rightarrow FeSO_4+SO_2\uparrow+2H_2O\left(3\right)\)
\(Theo.pt\left(2\right):n_{H_2S}=n_{FeS}=0,1\left(mol\right)\\ Theo.pt\left(3\right):n_{SO_2}=n_{Fe}=0,1\left(mol\right)\\ \%V_{H_2S}=\dfrac{0,1}{0,1+0,1}=50\%\\ \%V_{SO_2}=100\%-50\%=50\%\)
\(Theo.pt\left(2\right):n_{H_2SO_4\left(2\right)}=n_{FeS}=0,1\left(mol\right)\\ Theo.pt\left(3\right):n_{H_2SO_4\left(3\right)}=2n_{Fe}=2.0,1=0,3\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,3}{0,2}=1,5M\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
\(n_S=\dfrac{4,8}{32}=0,15mol\)
\(Fe+S\underrightarrow{t^o}FeS\)
0,1 0,15 0,1
\(FeS+2HCl\rightarrow FeCl_2+H_2S\uparrow\)
0,1 0,1 0,1
\(V_Y=0,1\cdot22,4=2,24l\)
\(m_Z=m_{FeCl_2}=0,1\cdot127=12,7g\)
Câu 30:
Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
Mg + H2SO4 ---> MgSO4 + H2
a--------------------------------->a
Zn + H2SO4 ---> ZnSO4 + H2
b--------------------------------->b
Hệ pt \(\left\{{}\begin{matrix}24a+65b=8,9\\a+b=0,2\end{matrix}\right.\Leftrightarrow a=b=0,1\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Zn}=0,1.65=6,5\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
gọi nMg : a ., nZn : b
=> 24a+65b = 8,9 (g)
pthh: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b b
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
a a
=> a+b = 0,2
=> \(\left\{{}\begin{matrix}24a+65b=8,9\\a+b=0,2\end{matrix}\right.\)
=> a = 0,1(mol) , b = 0,1 (mol)
=> \(m_{Mg}=0,1.24=2,4\left(g\right)m_{Zn}=0,1.65=6,5\left(g\right)\)
C23:
từ h/c ACO3 và BCO3
=> A , B có hóa trị II .
PTHH:
\(ACO_3+2HCl\rightarrow ACl_2+H_2O+CO_2\)
\(BCO_3+2HCl\rightarrow BCl_2+H_2O+CO_2\)
\(nCO_2=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
= > \(2nHCl=nCO_2\)
=> \(nHCl=0,4:2=0,2\left(mol\right)\)
= > \(VHCl=\dfrac{nHCl}{CM_{HCl}}=\dfrac{0,2}{1}=0,2\left(l\right)=200ml\)
chọn B
C24:
\(HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\)
1 1 1 1 (mol)
0,2 0,2 0,2 0,2 (mol)
kết tủa trắng là AgCl
= > mAgCl = 28,7 (g)
=>\(nAgCl=\dfrac{28,7}{143,5}=0,2\left(mol\right)\)
\(mHCl=0,2.36,5=7,3\left(g\right)\)
\(mAgNO_3=0,2.170=34\left(g\right)\)
\(mHNO_3=0,2.63=12,6\left(g\right)\)
mdd = mdd HCl + mAgNO3 + mHNO3 - mAgCl
= 20 + 34 + 12,6 - 28,7 = 37,9 (g)
\(C\%_{ddHCl}=\dfrac{7,3.100}{37,9}=19,26\%\)