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![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi kim loại hóa trị II cần tìm là A.
\(A+Cl_2\underrightarrow{to}ACl_2\\ ACl_2+2AgNO_3\rightarrow A\left(NO_3\right)_2+2AgCl\\ m_{\downarrow}=m_{AgCl}=86,1\left(g\right)\\ n_{AgCl}=\dfrac{86,1}{143,5}.100=0,6\left(mol\right)\\ n_A=n_{ACl_2}=n_{Cl_2}=n_{A\left(NO_3\right)_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ M_A=\dfrac{41,1}{0,3}=137\left(\dfrac{g}{mol}\right)\\ \rightarrow A:Bari\left(Ba=137\right)\\ b.V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c.m_{muối}=m_{Ba\left(NO_3\right)_2}=0,3.261=78,3\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.2B+6HCl\rightarrow2BCl_3+3H_2\\ H_2+CuO\underrightarrow{to}Cu+H_2O\\ n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\\ n_{H_2}=n_{Cu}=0,5\left(mol\right)\\ b.V_{X\left(đktc\right)}=V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(lít\right)\\ c.n_B=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\\ \rightarrow M_B=\dfrac{9}{\dfrac{1}{3}}=27\left(\dfrac{g}{mol}\right)\\ \rightarrow B:Nhôm\left(Al=27\right)\)
Giải thích các bước giải:
Gọi nFe = a mol ; nCu = b mol
⇒ 56a + 64b = 40 (1)
PTHH :
2Fe + 6H2SO4 → Fe2(SO4)3 + 3SO2 + 6H2O
a 3a 1,5a (mol)
Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
b 2b b (mol)
⇒ nSO2 = 1,5a + b =
15,68
22,4
= 0,7 (2)
Từ (1) và (2) suy ra : a = 0,12 ; b = 0,52
có : %mFe =
0,12.56
40
.100% = 16,8%
⇒ %mCu = 100% - 16,8% = 83,2%
Theo PT , có nH2SO4 = 3a + 2b = 0,12.3 + 0,52.2 = 1,4 mol
⇒ mH2SO4 = 1,4.98 = 137,2 gam
⇒ m dung dịch H2SO4 =
137,2
98
= 140 gam
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
A + 2HCl \(\rightarrow\)ACl2 + H2 (1)
ACl2 + 2AgNO3 \(\rightarrow\)A(NO3)2 + 2AgCl (2)
nAgCl=\(\dfrac{57,4}{143,5}=0,4\left(mol\right)\)
Theo PTHH 2 ta có:
nACl2=\(\dfrac{1}{2}\)nAgCl=0,2(mol)
Theo PTHH 1 ta có:
nACl2=nA=0,2(mol)
nHCl=2nACl2=0,4(mol)
MA=\(\dfrac{13}{0,2}=65\)
V dd HCl=0,4:2=0,2(lít)
Bài 2 :
Ta có PTHH :
(1) \(2B+6HCl->2BCl3+3H2\uparrow\)
1/3mol.......................................0,5mol
(2) \(H2+CuO-^{t0}->Cu+H2O\)
0,5mol..............................0,5mol
Theo đề bài ta có : nCu = \(\dfrac{32}{64}=0,5\left(mol\right)\)
VH2(ĐKTc) = 0,5.22,4 = 11,2( l)
MB = \(\dfrac{9}{\dfrac{1}{3}}=27\left(nh\text{ận}\right)\left(Al=27\right)\)
Vậy B là nhôm Al
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
$R + 2HCl \to RCl_2 + H_2$
$2B + 6HCl \to 2BCl_3 + 3H_2$
$n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$n_{HCl} = 2n_{H_2} = 0,5(mol)$
Bảo toàn khối lượng : $m_{muối} = m_{kl} + m_{HCl} - m_{H_2}$
$= 9,2 + 0,5.36,5 - 0,25.2 = 26,95(gam)$
b) $V_{dd\ HCl} = \dfrac{0,5}{2} = 0,25(lít)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt kim loại hóa trị II cần tìm là A.
\(n_{Ag}=\dfrac{75,6}{108}=0,7\left(mol\right)\\ A+2AgNO_3\rightarrow A\left(NO_3\right)_2+2Ag\\ n_A=\dfrac{n_{Ag}}{2}=\dfrac{0,7}{2}=0,35\left(mol\right)\\ M_A=\dfrac{m_A}{n_A}=\dfrac{19,6}{0,35}=56\left(\dfrac{g}{mol}\right)\)
Vậy kim loại A (II) cần tìm là sắt (Fe=56)
b)
\(n_{AgNO_3}=n_{Ag}=0,7\left(mol\right)\\ C_{MddAgNO_3}=\dfrac{0,7}{0,14}=5\left(M\right)\)
c)
\(V_{ddsau}=V_{ddAgNO_3}=0,14\left(l\right)\\ C_{MddFe\left(NO_3\right)_2}=\dfrac{0,35}{0,14}=2,5\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Mg + H2SO4 --> MgSO4 + H2
b) \(n_{Mg}=\dfrac{3}{24}=0,125\left(mol\right)\)
PTHH: Mg + H2SO4 --> MgSO4 + H2
0,125->0,125-->0,125-->0,125
=> VH2 = 0,125.22,4 = 2,8 (l)
\(V_{dd.H_2SO_4}=\dfrac{0,125}{2}=0,0625\left(l\right)\)
c) Sản phẩm là Magie sunfat và khí hidro
\(m_{MgSO_4}=0,125.120=15\left(g\right)\)
mH2 = 0,125.2 = 0,25 (g)
d)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,125}{1}\) => Hiệu suất tính theo H2
Gọi số mol CuO bị khử là a (mol)
PTHH: CuO + H2 --to--> Cu + H2O
a--->a-------->a
=> 16 - 80a + 64a = 14,4
=> a = 0,1 (mol)
=> nH2(pư) = 0,1 (mol)
=> \(H=\dfrac{0,1}{0,125}.100\%=80\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ a............3a.......a.........1,5a\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.........2b........b.........b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}27a+56b=5,5\\1,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,5}.100\approx49,091\%\\\%m_{Fe}=\dfrac{0,05.56}{5,5}.100\approx50,909\%\end{matrix}\right.\\ b.C_{MddHCl}=\dfrac{3a+2b}{0,5}=\dfrac{3.0,1+2.0,05}{0,5}=0,8\left(M\right)\)
\(a.A+2HCl\rightarrow ACl_2+H_2\\ ACl_2+2AgNO_3\rightarrow A\left(NO_3\right)_2+2AgCl\downarrow\\ n_{AgCl\downarrow}=\dfrac{57,4}{143,5}=0,4\left(mol\right)\\ n_A=n_{ACl_2}=\dfrac{n_{AgCl}}{2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ b.M_A=\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\\ \rightarrow A:Kẽm\left(Zn=65\right)\\ c.n_{HCl}=2.n_A=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)