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\(n_{MnO_2}=\dfrac{17,4}{87}=0,2\left(mol\right)\\ PTHH:MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ n_{Cl_2\left(TT\right)}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\\ n_{Cl_2\left(LT\right)}=n_{MnO_2}=0,2\left(mol\right)\\ \Rightarrow H=\dfrac{n_{Cl_2\left(TT\right)}}{n_{Cl_2\left(LT\right)}}.100\%=\dfrac{0,16}{0,2}.100=80\%\)
- PT: a, \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)
\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\) (2)
- Ta có: \(n_{HCl\left(1\right)}=n_{HCl\left(2\right)}=0,2.2=0,4\left(mol\right)\)
Theo PT (1): \(n_{Cl_2}=\dfrac{5}{16}n_{HCl\left(1\right)}=0,125\left(mol\right)\Rightarrow V_1=0,125.22,4=2,8\left(l\right)\)
(2): \(n_{Cl_2\left(2\right)}=\dfrac{1}{4}n_{HCl\left(2\right)}=0,1\left(mol\right)\Rightarrow V_2=0,1.22,4=2,24\left(l\right)\)
Đáp án C
MnO2 + 4HCl →MnCl2 + 2H2O + Cl2
0,1 →0,1 (mol)
Do H% = 85% => 
= 0,085 (mol)
V = 0,085.22,4 = 1,904 (lít)
\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \left(mol\right).....0,15\rightarrow................0,15\\ V_{H_2}=0,15.22,4=3,36\left(l\right)\)
1)
Fe + 2HCl --> FeCl2 + H2
Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
2)
- Xét TN1:
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15<------------------0,15
=> mFe = 0,15.56 = 8,4 (g)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{8,4}{14,8}.100\%=56,757\%\\\%m_{Cu}=100\%-56,757\%=43,243\%\end{matrix}\right.\)
3)
- Xét TN2:
\(n_{Cu}=\dfrac{29,6.43,243\%}{64}=0,2\left(mol\right)\)
PTHH: Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
0,2-------------------------->0,2
=> V = 0,2.22,4 = 4,48 (l)
Anh nghĩ 3.65 (g) là khối lượng của HCl luôn ấy em
\(n_{HCl}=\dfrac{3.65}{36.5}=0.1\left(mol\right)\)
\(MnO_2+4HCl_{\left(đ\right)}\rightarrow MnCl_2+Cl_2+2H_2O\)
\(............0.1.............0.025\)
\(V_{Cl_2}=0.025\cdot22.4=0.56\left(l\right)\)
*Chắc là 3,65 gam HCl
PTHH: \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2\uparrow+H_2O\)
Theo PTHH: \(n_{Cl_2}=\dfrac{1}{4}n_{HCl}=\dfrac{1}{4}\cdot\dfrac{3,65}{36,5}=0,025\left(mol\right)\)
\(\Rightarrow V_{Cl_2}=0,025\cdot22,4=0,56\left(l\right)\)