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a) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15\cdot56=8,4\left(g\right)\)
c) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(V_{H_2SO_4}=50ml=0,05\left(l\right)\)
\(\Rightarrow C_{M,H_2SO_4}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_4}}=\dfrac{0,15}{0,05}=3M\)
a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
\(n_{Fe}=n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
`a)PTHH:`
`Fe + 2HCl -> FeCl_2 + H_2`
`0,4` `0,4` `0,4`
`n_[H_2] = [ 8,96 ] / [ 22,4 ] = 0,4 (mol)`
`b) m_[Fe] = 0,4 . 56 = 22,4 (g)`
`c) m_[FeCl_2] = 0,4 . 127 = 50,8 (g)`
Fe+2HCl->FeCl2+H2
0,3----0,6--0,3------0,3
n H2=0,3 mol
=>m Fe=0,3.56=16.8g
=>CM=0,6\0,1=6M
CuO+H2-tO>Cu+H2O
0.3---------------0,3
=>m Cu=0,3.64=19,2g
Sửa đề : 11.2 g sắt
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0.2....0.2.................0.2\)
\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)
Theo phương trình phản ứng ta có:
n F e = n H 2 = 0,075 mol
n H 2 S O 4 = 0,075 mol (mà H 2 S O 4 đề cho là 0,2 mol nên H 2 S O 4 dư)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,075\left(mol\right)\Rightarrow m_{Fe}=0,075.56=4,2\left(g\right)\)
c, \(n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,075\left(mol\right)\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2-0,075=0,125\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,125.98=12,25\left(g\right)\)
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0.25....0.25.......................0.25\)
\(m_{Fe}=0.25\cdot56=14\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.25}{0.1}=2.5\left(M\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{9,9161}{24,79}\approx0,4\left(mol\right)\\ b,m_{Fe}\approx0,4.56\approx22,4\left(g\right)\\ c,n_{HCl}\approx0,4.2\approx0,8\left(mol\right)\\ C_{MddHCl}\approx\dfrac{0,8}{0,25}\approx3,2\left(M\right)\)
\(a)Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{9,9161}{24,79}\approx0,4mol\\ n_{Fe}=n_{H_2}=0,4mol\\ m_{Fe}=0,4.56=22,4g\\ c)n_{HCl}=2n_{H_2}=2.0,4=0,8mol\\ 250ml=0,25l\\ C_{M_{HCl}}=\dfrac{0,8}{0,25}=3,2M\)