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Đặt \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\Rightarrow65x+24y=8,9\left(1\right)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow x+y=0,2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}65x+24y=8,9\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\\ \Rightarrow\%_{Zn}=\dfrac{0,1\cdot65}{8,9}\cdot100\%\approx73\%\\ \Rightarrow\%_{Mg}=100\%-73\%=27\%\)
\(n_{HCl}=2x+2y=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{14,6\cdot100\%}{14,6\%}=100\left(g\right)\)
\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\
Fe+2HCl\rightarrow FeCl_2+H_2\\
n_{Fe}=n_{H_2}=0,2mol\\
m_{Fe}=0,2.56=11,2g\\
m_{Cu}=25-11,2=13,8g\\
b,\%m_{Fe}=\dfrac{11,2}{25}\cdot100=44,8\%\\
\%m_{Cu}=100-44,8=55,2\%\)
c, Gọi CTHH của sắt là \(Fe_xO_y\)
\(Fe_xO_y+yH_2\xrightarrow[t^0]{}xFe+yH_2O\\ \Rightarrow n_{Fe_xO_y}=n_{H_2}:y\\ \Leftrightarrow\dfrac{11,6}{56x+16y}=\dfrac{0,2}{y}\\ \Leftrightarrow11,6y=11,2x+3,2y\\ \Leftrightarrow11,6y-3,2y=11,2x\\ \Leftrightarrow8,4y=11,2x\\ \Leftrightarrow\dfrac{x}{y}=\dfrac{8,4}{11,2}=\dfrac{3}{4}\\ \Rightarrow x=3;y=4\\ \Rightarrow CTHH:Fe_3O_4\)
\(A.Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ B.n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,05 0,05 0,05 0,05
\(\%m_{Mg}=\dfrac{0,05.24}{6,4}\cdot100=18,75\%\\ \%m_{Cu}=100-18,75=81,25\%\\ C.m_{ddH_2SO_4}=\dfrac{0,05.98}{20}\cdot100=24,5g\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Theo Pt : \(n_{Mg}=n_{H2SO4}=n_{MgSO4}=n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
b) \(\%m_{Mg}=\dfrac{0,05.24}{6,4}.100\%=18,75\%\)
\(\%m_{Cu}=100\%-18,75\%=81,25\%\)
c) \(m_{H2SO4}=0,05.98=4,9\left(g\right)\)
\(\Rightarrow m_{ddH2SO4}=\dfrac{4.100\%}{20\%}=20\left(g\right)\)
Chúc bạn học tốt
a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
X là khí Hidro
b) Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Al}=0,2mol\)
\(\Rightarrow\%m_{Al}=\dfrac{0,2\cdot27}{8,64}\cdot100\%=62,5\%\) \(\Rightarrow\%m_{Cu}=37,5\%\)
c) Theo PTHH: \(n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow V_{HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl -->ZnCl2 + H2
____0,2<----------------------0,2
=> mZn = 0,2.65 = 13 (g)
mCu = mrắn không tan = 19,5 (g)
\(\left\{{}\begin{matrix}\%Zn=\dfrac{13}{13+19,5}.100\%=40\%\\\%Cu=\dfrac{19,5}{13+19,5}.100\%=60\%\end{matrix}\right.\)
`n_(H_2)=4,48/22,4=0,2 (mol)`
Ta có PTHH: `Zn+2HCl --> ZnCl_2 +H_2`
Theo PT: `1`--------------------------------`1`
Theo đề: `0,2`------------------------------`0,2`
`m_(Zn)=0,2.65=13(g)`
Vì `Cu` không phản ứng với `HCl` nên `m_(chất rắn không tan)=m_(Cu)=19,5(gam)`
`%Zn=13/(13+19,5) .100%=40%`
`%Cu=100%-40%=60%`
a) nNaOH = 0,6.1 = 0,6 (mol)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,6----->0,6
=> mCH3COOH = 0,6.60 = 36 (g)
=> mC2H5OH = 45,2 - 36 = 9,2 (g)
b) \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
PTHH: 2CH3COOH + 2Na --> 2CH3COONa + H2
0,6---------------------------->0,3
2C2H5OH + 2Na --> 2C2H5ONa + H2
0,2--------------------------->0,1
=> V = (0,3 + 0,1).22,4 = 8,96 (l)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH:
\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)
\(Cu+HCl--\times-->\)
b. Theo PT(1): \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\Rightarrow\%_{m_{Mg}}=\dfrac{4,8}{11,2}.100\%=42,9\%\)
\(\%_{m_{Cu}}=100\%-42,9\%=57,1\%\)
c. Theo PT(1): \(n_{HCl}=2.n_{H_2}=2.0,2=0,4\left(mol\right)\)
PTHH: \(NaOH+HCl--->NaCl+H_2O\left(2\right)\)
Theo PT(2): \(n_{NaOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)
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