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a)
Gọi số mol Fe, Zn là a, b (mol)
=> 56a + 65b = 5,49 (1)
\(n_{H_2}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\)
nHCl(bđ) = 0,3.1 = 0,3 (mol)
Do nHCl(bđ) > 2.nH2 => HCl dư
PTHH: Fe + 2HCl --> FeCl2 + H2
a-------------->a------>a
Zn + 2HCl --> ZnCl2 + H2
b-------------->b------>b
=> a + b = 0,09 (2)
(1)(2) => a = 0,04 (mol); b = 0,05 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,04.56}{5,49}.100\%=40,8\%\\\%m_{Zn}=\dfrac{0,05.65}{5,49}.100\%=59,2\%\end{matrix}\right.\)
b)
mFeCl2 = 0,04.127 = 5,08 (g)
mZnCl2 = 0,05.136 = 6,8 (g)
=> mmuối = 5,08 + 6,8 = 11,88 (g)
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Ta có: \(\left\{{}\begin{matrix}p+e+n=52\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=17\\n=18\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Khúc đầu bt lm vầy thoi còn đoạn sau sợ làm quá nó lố lăng nên để bn lm:3
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2 :
a) PTHH : \(2Al+3S-->Al_2S_3\)
\(Zn+S-->ZnS\)
mS = 29,55 - 15,15 = 14,4 (g)
b) nS = 14,4/32 = 0,45 (mol)
PTHH : \(Al_2S_3+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2S\)
\(ZnS+H_2SO_4-->ZnSO_4+H_2S\)
\(H_2S+2AgNO_3-->Ag_2S+2HNO_3\)
Bảo toàn S : nAg2S = nH2S = nS = 0,45 (mol)
=> mktAg2S = 111,6 (g)
c) PTHH : \(H_2S+CuCl_2-->CuS+2HCl\)
nH2S = 0,45/2 = 0,225 (mol)
Theo pthh : nCuS = nH2S = 0,225 (mol)
=> mCuS = 0,225.96 = 21,6 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Gọi số mol Fe, Al là a, b (mol)
=> 56a + 27b = 8,94 (1)
\(n_{H_2}=\dfrac{0,36}{2}=0,18\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
a------------->a----->a
2Al + 6HCl --> 2AlCl3 + 3H2
b---------------->b------>1,5b
=> a + 1,5b = 0,18 (2)
(1)(2) => a = 0,15 (mol); b = 0,02 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{8,94}.100\%=93,96\%\\\%m_{Al}=\dfrac{0,02.27}{8,94}.100\%=6,04\%\end{matrix}\right.\)
b)
mFeCl2 = 0,15.127 = 19,05 (g)
mAlCl3 = 0,02.133,5 = 2,67 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi số mol Mg, Zn là a, b (mol)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a---------------->a----->a
Zn + 2HCl --> ZnCl2 + H2
b--------------->b---->b
=> a + b = 0,3 (1)
Và 95a + 136b = 36,7 (2)
(1)(2) => a = 0,1 (mol); b = 0,2 (mol)
\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{0,1.24+0,2.65}.100\%=15,6\%\\\%m_{Zn}=\dfrac{0,2.65}{0,1.24+0,2.65}.100\%=84,4\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Hiện tượng: mất dần màu vàng lục của Clo, có khí không màu thoát ra
\(K_2CO_3+Cl_2\rightarrow KCl+KClO+CO_2\uparrow\)
b, \(F_2+H_2\xrightarrow[\text{nhiệt độ âm}]{\text{bóng tối}}2HF\)
\(Cl_2+H_2\underrightarrow{as}2HCl\)
\(Br_2+H_2\underrightarrow{t^o}2HBr\\ I_2+H_2\underrightarrow{t^o,xt}2HI\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{HCl}=0,1.1,5=0,15\left(mol\right);n_{H_2}=\dfrac{1,288}{22,4}=0,0575\left(mol\right)\)
Ta có: \(\dfrac{0,15}{8}>\dfrac{0,0575}{4}\) ⇒ HCl dư
⇒ tính số mol theo H2
Fe + 2HCl ----> FeCl2 + H2
x x x x
2Al + 6HCl ----> 2AlCl3 + 3H2
y 3y 2y 1,5y
Ta có: hệ pt: \(\left\{{}\begin{matrix}56x+27y=1,795\\x+1,5y=0,0575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,025\left(mol\right)\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,02.56.100\%}{1,795}=62,4\%;\%m_{Al}=100\%-62,4\%=37,6\%\)
b, Trong dd X có HCl dư, FeCl2 và AlCl3
\(n_{HCldư}=0,15-0,02-3.0,025=0,055\left(mol\right)\)
HCl + KOH ----> KCl + H2O
0,055 0,055
2KOH + FeCl2 ----> Fe(OH)2 + 2KCl
0,04 0,02
3KOH + AlCl3 -----> Al(OH)3 + 3KCl
0,15 0,05
\(n_{KOH}=0,055+0,04+0,15=0,245\left(mol\right)\)
\(\Rightarrow V_{KOH}=\dfrac{0,245}{1}=0,245\left(l\right)=245\left(ml\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
a a a
2Al + 6HCl ---> 2AlCl3 + 3H2
b b 1,5b
\(n_{H_2}=\dfrac{1,228}{22,4}=0,0575\left(mol\right)\)
Hệ pt \(\left\{{}\begin{matrix}56a+27b=1,795\\a+1,5b=0,0575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,02\left(mol\right)\\b=0,025\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,02.56=1,12\left(g\right)\\\%m_{Al}=0,025.27=0,675\left(g\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{1,12}{1,795}=62,4\%\\\%m_{Al}=100\%-62,4\%=37,6\%\end{matrix}\right.\)
PTHH:
FeCl2 + 2KOH ---> Fe(OH)2 + 2KCl
0,02 0,04 0,02
AlCl3 + 3KOH ---> Al(OH)3 + 3KCl
0,025 0,075 0,025
\(\rightarrow V_{ddKOH}=\dfrac{0,075+0,04}{1}=0,115\left(l\right)\)
a)
\(n_{H_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
PTHH: 2R + 2HCl --> 2RCl + H2
1,5<---------------------0,75
=> \(M_R=\dfrac{34,5}{1,5}=23\left(Na\right)\)
b)
PTHH: 2Na + 2HCl --> 2NaCl + H2
1,5<------1,5<--0,75
=> mHCl = 1,5.36,5 = 54,75(g)
=> \(m_{ddHCl}=\dfrac{54,75.100}{20}=273,75\left(g\right)\)
c) mNaCl = 1,5.58,5 = 87,75(g)
mdd sau pư = 34,5 + 273,75 - 0,75.2 = 306,75(g)
=> \(C\%=\dfrac{87,75}{306,75}.100\%=28,6\%\)
Ta có kim loại kiềm có hóa trị II, nên ta có PT:
\(R+2HCl--->RCl_2+H_2\uparrow\left(1\right)\)
a. Ta có: \(n_{H_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
Theo PT(1): \(n_R=n_{H_2}=0,75\left(mol\right)\)
\(\Rightarrow M_R=\dfrac{34,5}{0,75}=46\left(\dfrac{g}{mol}\right)\)
Không có kim loại nào có M = 46 (g/mol)
(Mik nghĩ là đề sai)