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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\a, PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\\ c,C_{MddH_2SO_4}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\
b)200ml=0,2l\\
n_{HCl}=0,2.1=0,2mol\\
n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\
V_{H_2}=0,1.24,79=2,479l\\
c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nAl = 5.4 / 27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2......0.6............0.2.......0.3
a) VH2 = 0.3 * 22.4 = 6.72 (l)
b) mAlCl3 = 0.2 * 133.5 = 26.7 (g)
c) VddHCl = 0.6 / 1.5 = 0.4 (l)
d) CMAlCl3 = 0.2 / 0.4 = 0.5 (M)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đổi 200 ml = 0,02 l
a) PTHH : Fe + HCl -> FeCl2 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(=>V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(C_{HCl}=\dfrac{n}{V}=\dfrac{0,1}{0,02}=5\left(M\right)\)
Đông Hải làm câu 1 rồi thì tui làm phần còn lại
Câu 2:
\(a,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ b,n_{CuO}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\\ Theo.PTHH:n_{Cu}=n_{H_2}=n_{CuO}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\\ c,m_{Cu}=n.M=0,1.64=6,4\left(g\right)\\ d,PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\left(2\right)\\ Theo.PTHH\left(2\right):n_{H_2O}=n_{H_2}=0,1\left(mol\right)\)
\(m_{H_2O}=n.M=0,1.18=1,8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)
\(n_{HCl}=2,5.0,2=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 < 0,5 ( mol )
0,2 0,4 0,2 0,2 ( mol )
Chất dư là HCl
\(m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65g\)
\(V_{H_2}=0,2.22,4=4,48l\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,2}=1M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{22,4}{56}=0,4mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,4 0,8 0,4
\(C_{M_{HCl}}=\dfrac{0,8}{0,4}=2M\)
\(V_{H_2}=0,4\cdot22,4=8,96l\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a.
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(V_{H_2}=24,79.0,2=4,958\left(l\right)\)
b.
\(n_{HCl}=2.n_{Fe}=0,4\left(mol\right)\\ CM_{HCl}=\dfrac{0,4}{0,2}=2M\)
tớ cảm ơnn