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\(n_{Zn}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+H_2\)
\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(1\right)\)
\(m_{Muối}=m_{ZnCl_2}+m_{AlCl_3}=136a+133.5b=40.3\left(g\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(m_{hh}=0.1\cdot65+0.2\cdot27=11.9\left(g\right)\)
\(\%Zn=\dfrac{0.1\cdot65}{11.9}\cdot100\%=54.62\%\)
\(\%Al=100-54.62=45.38\%\)
a) 2Zn+O2--->2ZnO
x----------0,5x
4Al+3O2--->2Al2O3
y-------0,75y
n O2=5,6/22,4=0,25(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}65x+27y=18,4\\0,5x+0,75y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
%m Zn =0,2.65/18,4.100%=71,43%
%m Al=100%-71,43=28,57%
b) Zn+2HCl--->ZnCl2+H2(1)
2Al+6HCl---->2AlCl3+3H2(2)
Nếu cho 9,2g hh X như trên thì n Zn=n Al=0,1(mol)
theo pthh1
n H2=n Zn=0,1(mol)
Theo pthh2
n H2=3/2n Al=0,15(mol)
Tổng n H2=0,25(mol)
V H2=0,25.22,4=5,6(l)
Al, Fe không tác dụng với H2SO4 đặc nguội
Rắn không tan ở TN2 là Cu
mCu = 6,4 (g)
=> \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
PTHH: Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
0,1-------------------------->0,1
=> V = 0,1.22,4 = 2,24 (l)
a) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Gọi số mol Zn, Al là a, b
=> 65a + 27b = 18,4 (1)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a----->2a------------->a
2Al + 6HCl --> 2AlCl3 + 3H2
b---->3b------------->1,5b
=> a + 1,5b = 0,5 (2)
(1)(2) => a = 0,2 ; b = 0,2
=> \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
b) nHCl(pư) = 2a + 3b = 1 (mol)
nHCl(dư) = 0,6.2 - 1 = 0,2 (mol)
PTHH: KOH + HCl --> KCl + H2O
0,2<----0,2
=> \(V=\dfrac{0,2}{1}=0,2\left(l\right)\)
\(n_{HCl}=0,6.2=1,2\left(mol\right)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ a,n_{HCl\left(dư\right)}=1,2-2.n_{H_2}=1,2-2.0,5=0,2\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:n_{Zn}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+27b=18,4\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Al}=0,2.27=5,4\left(g\right)\\ b,KOH+HCl_{dư}\rightarrow KCl+H_2O\\ n_{KOH}=n_{HCl\left(dư\right)}=0,2\left(mol\right)\\ \Rightarrow V=V_{ddKOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)
a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)
Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)
Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Zn}:b\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}56a+65b=18,6\\127a+136b=39,9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,1.56}{18,6}.100\%=30,11\%\)
\(\%m_{Fe}=100\%-30,11\%=69,89\%\)
b, \(n_{HCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(V_{dd\left(HCl\right)}=\frac{0,6}{4}=0,15\left(l\right)=150\left(ml\right)\)
\(m_{dd\left(HCl\right)}=150.1,1=165\left(g\right)\)
c,\(ZnCl_2+2AgNO_3\rightarrow Zn\left(NO_3\right)_2+2AgCl\)
\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_3+2AgCl\)
\(n_{AgCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,6.143,5=86,1\left(g\right)\)
Câu 1:
Đặt \(n_{Fe}=x\left(mol\right);n_{Al}=y\left(mol\right)\)
\(n_{HCl}=0,4\left(mol\right)\)
\(Fe^o\rightarrow Fe^{+2}+2e\)
x_____________2x_(mol)
\(Al^o\rightarrow Al^{+3}+3e\)
y____________3y_(mol)
\(2H^-\rightarrow H_2^o+2e\)
0,8_____0,4____0,8_(mol)
\(BTe:2x+3y=0,8\)
Theo đề ta có hệ: \(\left\{{}\begin{matrix}56x+27y=11\\2x+3y=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{11}.100\%=51\left(\%\right)\\\%m_{Al}=100-51=49\left(\%\right)\end{matrix}\right.\)
\(BTNT:\Rightarrow\left\{{}\begin{matrix}n_{FeCl_2}=0,1\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\)
\(m_{hh}=0,1.127+133,5.0,2=39,4\left(g\right)\)
\(m_{ddHCl}=\frac{36,5.0,8.100}{7,3}=400\left(g\right)\)
Câu 2:
\(n_{H_2}=0,15\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
(mol)____0,1____0,3______0,1______0,15__
\(\%m_{Al_2O_3}=\frac{7,8-27.0,1}{7,8}.100\%=65,4\left(\%\right)\)
Câu 3:
\(n_{H_2}=0,1\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
(mol)_____0,1__________________0,1__
\(\%m_{ZnO}=\frac{10,55-0,1.65}{10,55}.100\%=38,4\left(\%\right)\)
a) 2Al+6HCl---.>2AlCl3+3H2
x-------------------------------1,5x
Zn+2HCl--->ZnCl2+H2
y-----------------------------y
n H2=5,6/22,4=0,25(mol)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}27x+65y=9,2\\1,5x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
%m Al=0,1.27/9,2.100%=29,35%
%m Zn=100%-29,35=70,65%
b)m muối=0,1.133,5+0,1.136=26,95(g
Bt lm phần c ko ạ