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a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{54}{27}=2\left(mol\right)\)
Theo PTHH: \(n_{H_2}=\dfrac{2.3}{2}=3\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=3.22,4=67,2l\)
b) \(2H_2+O_2\rightarrow2H_2O\)
\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{30}{32}=0,94\left(mol\right)\)
Theo PTHH: \(n_{H_2O}=\dfrac{0,94.2}{1}=1,88\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=1,88.18=33,84\left(g\right)\)
a, Cu không tác dụng với dd HCl.
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=1\left(mol\right)\Rightarrow m_{Zn}=1.65=65\left(g\right)\)
\(\Rightarrow m_{Cu}=80,5-65=15,5\left(g\right)\)
Bài 6:
\(2M+3Cl_2\rightarrow2MCl_3\\ m_{Cl_2}=m_{MCl_3}-m_M=32,5-11,2=21,3\left(g\right)\\ n_{Cl_2}=\dfrac{21,3}{71}=0,3\left(mol\right)\\ n_M=\dfrac{2}{3}.n_{Cl_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ M_M=\dfrac{m_M}{n_M}=\dfrac{11,2}{0,2}=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow M\left(III\right):Sắt\left(Fe=56\right)\)
Bài 5:
\(KL:X\left(II\right)\\ X+2HCl\rightarrow XCl_2+H_2\\ n_{HCl}=0,4.0,3=0,12\left(mol\right)\\ n_X=n_{XCl_2}=n_{H_2}=\dfrac{0,12}{2}=0,06\left(mol\right)\\ a,M_X=\dfrac{3,36}{0,06}=56\left(\dfrac{g}{mol}\right)\Rightarrow X\left(II\right):Sắt\left(Fe=56\right)\\ b,V_{ddFeCl_2}=V_{ddHCl}=0,4\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,06}{0,4}=0,15\left(M\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,05 0,15
\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)
\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)
a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)
\(\%m_{Ag}=100\%-64,28\%=35,72\%\)
b)\(m_{muối}=0,05\cdot342=17,1g\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(0.2.................0.1...........0.1...........0.1\)
Bảo toàn khối lượng :
\(m_{O_2}=37.92-34.72=3.2\left(g\right)\)
\(n_{O_2}=\dfrac{3.2}{32}=0.1\left(mol\right)\)
\(m_{KMnO_4\left(dư\right)}=37.92-0.2\cdot158=6.32\left(g\right)\)
\(m_{K_2MnO_4}=0.1\cdot197=19.7\left(g\right)\)
\(m_{MnO_2}=0.1\cdot87=8.7\left(g\right)\)
\(b.\)
Để hỗn hợp nổ mạnh nhất thì H2 phản ứng với O2 theo tỉ lệ 2 : 1
\(2H_2+O_2\underrightarrow{^{^{t^0}}}2H_2O\)
\(0.2......0.1\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{2}{15}...........................0.2\)
\(m_{Al}=\dfrac{2}{15}\cdot27=3.6\left(g\right)\)
2Al+6HCl→2AlCl3+3H2
Zn+2HCl→ZnCl2+H2
2Al+6H2SO4→Al2(SO4)3+3SO2+6H2O
Zn+2H2SO4→ZnSO4+SO2+2H2O
Cu+2H2SO4→CuSO4+SO2+2H2O
nH2=0,3mol
nCu=0,15mol
Gọi a và b lần lượt là số mol của Al và Zn
27a+65b=17,25
3\2a+b=0,3
=> a=0,03, b=0,25
→nAl=0,03mol→mAl=1,62g
→nZn=0,25mol→mZn=32,5g
b)nHCl=3nAl+2nZn=0,59mol
→VHCl=0,592=0,295 l
c)
nAl2(SO4)3=1\2nAl=0,015mol
→mAl2(SO4)3=5,13g
nZnSO4=nZn=0,25mol
→mZnSO4=40,25g
nCuSO4=nCu=0,15mol
→mCuSO4=24g