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a, \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(n_{Fe}=n_{H_2}=0,6\left(mol\right)\Rightarrow m_{Fe}=0,6.56=33,6\left(g\right)\)
c, \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,6}{1}\), ta được CuO pư hết.
a, nH2 = V/22,4 = 13,44/22,4 =0.6 (mol)
Fe + 2HCl \(\rightarrow \) FeCl2 + H2
TLM : 1 2 1 1
Đề cho: 0,6<--1,2<----------- 0,6 (mol)
mHCl = n . M = 1,2 . 36,5 = 43,8 (g)
mFe= n . M = 0,6 . 56 =33,6 (g)
c, nCuO = \(\dfrac{16}{80}\)= 0,2 (mol)
CuO + H2 \(\rightarrow \) Cu + H2O
TLM: 1 1 1 1
Vì \(\dfrac{nH_2}{1}\)= 0,6 < \(\dfrac{n_{CuO}}{1}\)= 0.2
=> CuO phản ứng hết.
\(a)\\ Fe + 2HCl \to FeCl_2 + H_2\)
b)
\(n_{Fe} = \dfrac{22,4}{56}= 0,4(mol)\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)
Ta thấy : \(n_{Fe} > n_{H_2}\) nên Fe dư.
Theo PTHH :
\(n_{Fe\ pư} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{Fe\ pư} = 0,3.56 = 16,8(gam)\)
c)
Ta có :
\(n_{FeCl_2} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{FeCl_2} = 0,3.127 = 38,1(gam)\)
a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
nFe = 16,8/56 = 0,3 (mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,3 ---> 0,6 ---> 0,3 ---> 0,3
VH2 = 0,3 . 24,79 = 7,437 (l)
mHCl = 0,6 . 36,5 = 21,9 (g)
PTHH: CuO + H2 -> (t°) Cu + H2O
Mol: 0,3 <--- 0,3 ---> 0,3
mCu = 0,3 . 64 = 19,2 (g)
mFe = 16,8: 56 =0,3(mol)
pthh : Fe + 2HCl --> FeCl2 + H2 (1)
0,3 ->0,6-----------------> 0,3 (mol)
=> VH2 (đkc) = 0,3 . 24,79 ( l)
=> mHCl = 0,6 . 35,5 = 21,9 (g)
pthh : CuO + H2 -t--> Cu+ H2O
0,3<-----0,3 (mol)
=>mCu = 0,3 . 64 = 19,2 (g)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)
a)
$Zn + 2HCl \to ZnCl_2 + H_2$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$
Theo PTHH : $n_{Zn} = n_{H_2} = 0,06(mol)$
$\Rightarrow n_{Fe_2O_3} = \dfrac{7,1-0,06.65}{160} = 0,02(mol)$
Theo PTHH : $n_{HCl} = 2n_{Zn} + 6n_{Fe_2O_3} = 0,24(mol)$
$m_{HCl} = 0,24.36,5 = 8,76(gam)$
b)
Gọi $n_{CuO} = x(mol) ; n_{Fe_3O_4} = y(mol) \Rightarrow 80a + 232y = 3,92(1)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$
Theo PTHH : $n_{H_2} =x + 4y = 0,06(2)$
Từ (1)(2) suy ra: x = 0,02; y = 0,01
$n_{Cu} = 0,02(mol) \Rightarrow m_{Cu} = 0,02.64 = 1,28(gam)$
$n_{Fe} = 0,01.3 = 0,03(mol) \Rightarrow m_{Fe} = 0,03.56 = 1,68(gam)$
a)
\(Zn + 2HCl \to ZnCl_2 + H_2\\ Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2\)
Theo PTHH : \(n_{Zn} = n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)
\(\Rightarrow n_{Fe_2O_3} = \dfrac{35,5-0,3.65}{160} = 0,1\\ \Rightarrow n_{HCl} = 2n_{Zn} + 6n_{Fe_2O_3} = 0,3.2 + 0,1.6 = 1,2(mol)\\ \Rightarrow m_{HCl} = 1,2.36,5 = 43,8(gam)\)
b)
\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\)
Gọi \(n_{CuO} = a;n_{Fe_2O_3} = b\)
\(\left\{{}\begin{matrix}80a+160b=19,6\\a+3b=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,135\\b=0,055\end{matrix}\right.\)
Vậy :
\(\left\{{}\begin{matrix}n_{Cu}=0,135\\n_{Fe}=0,055.2=0,11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,135.64=8,64\left(gam\right)\\m_{Fe}=0,11.56=6,16\left(gam\right)\end{matrix}\right.\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{65}{65}=1\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=1.136=136\left(g\right)\\ c.n_{H_2}=n_{Zn}=1\left(mol\right)\\ \Rightarrow V_{H_2}=1.22,4=22,4\left(l\right)\)
\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(1\right)\\ b,Theo.pt\left(1\right):n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ m_{Zn}=0,3.65=19,5\left(g\right)\\ Theo.pt\left(1\right):n_{HCl}=2n_{H_2}=2.0,3=0,6\left(mol\right)\\ m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(c,m_{Fe}=94,03\%.16,08\approx11,2\left(g\right)\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{H_2}=n_{O\left(trong.Fe_xO_y\right)}=0,3\left(mol\right)\\ CTPT:Fe_xO_y\\ \Rightarrow x:y=0,2:0,3=2:3\\ CTPT:Fe_2O_3\)
Thầy em ra là Fe3O4 ạ