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Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\
b)200ml=0,2l\\
n_{HCl}=0,2.1=0,2mol\\
n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\
V_{H_2}=0,1.24,79=2,479l\\
c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(n_{H_2SO_4}=0.3\cdot1=0.3\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.1........0.1...........0.1.......0.1\)
\(\Rightarrow H_2SO_4dư\)
\(n_{H_2SO_4\left(dư\right)}=0.3-0.1=0.2\left(mol\right)\)
\(n_{ZnSO_4}=n_{H_2}=0.1\left(mol\right)\)
\(C_{M_{ZnSO_4}}=\dfrac{0.1}{0.3}=0.33\left(M\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2}{0.3}=0.66\left(M\right)\)
\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{Zn} = \dfrac{6,5}{65} = 0,1 < n_{H_2SO_4} =0,3 \to H_2SO_4\ dư\\ n_{H_2SO_4\ pư} = n_{ZnSO_4} = n_{Zn} = 0,1(mol)\\ n_{H_2SO_4\ dư} = 0,3 - 0,1 = 0,2(mol)\\ c) C_{M_{ZnSO_4}} = \dfrac{0,1}{0,3} = 0,33M\\ C_{M_{H_2SO_4}} = \dfrac{0,2}{0,3} = 0,67M\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\\ b,C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
a) \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,05->0,1----->0,05---->0,05
`=> V_{ddHCl} = (0,1)/2 = 0,05 (l)`
b) `V_{H_2} = 0,05.22,4 = 1,12 (l)`
c) `C_{M(FeCl_2)} = (0,05)/(0,05) = 1M`
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) HCl còn dư, Magie p/ứ hết
\(\Rightarrow n_{Mg}=n_{MgCl_2}=n_{H_2}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Mg}+m_{ddHCl}-m_{H_2}=111,7\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{9,5}{111,7}\cdot100\%\approx8,5\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{111,7}\cdot100\%\approx3,27\%\end{matrix}\right.\)
\(a)Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b)n_{Mg}=\dfrac{3}{24}=0,125mol\\ n_{HCl}=0,1.1=0,1mol\\ \Rightarrow\dfrac{0,125}{1}>\dfrac{0,1}{2}\Rightarrow Mg.dư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,1}{2}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5M\)
Đoạn xét tỉ lệ phải là \(\dfrac{0,125}{1}>\dfrac{0,1}{2}\) em nhé.