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a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c, \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{2}{15}.56=\dfrac{112}{15}\left(g\right)\)
PT: Fe2O3+3H2to→2Fe+3H2O
CuO+H2to→Cu+H2O
a, Ta có: mFe2O3=20.60%=12(g)
⇒nFe2O3=\(\dfrac{12}{160}\)=0,075(mol
mCuO=20−12=8(g
⇒nCuO=\(\dfrac{8}{80}\)=0,1(mol)
Theo pT:
nFe=2nFe2O3=0,15(mol)
nCu=nCuO=0,1(mol)
⇒mFe=0,15.56=8,4(g)
mCu=0,1.64=6,4(g)
b, Theo PT: nH2=3nFe2O3+nCuO=0,325(mol)
⇒VH2=0,325.22,4=7,28(l)
c. Zn+2HCl->ZnCl2+H2
0,65----------0,325
=>m HCl=0,65.36,5=23,725g
\(n_{Fe_2O_3}=\dfrac{32}{160}=0.2\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^o}}}2Fe+3H_2O\)
\(0.2........0.6........0.4........0.6\)
\(V_{H_2}=0.6\cdot22.4=13.44\left(l\right)\)
\(m_{Fe}=0.4\cdot56=22.4\left(g\right)\)
Số phân tử H2O là : \(0.6\cdot6\cdot10^{23}=3.6\cdot10^{23}\left(pt\right)\)
a.b.
\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 ( mol )
\(V_{H_2}=0,05.22,4=1,12l\)
\(m_{HCl}=0,1.36,5=3,65g\)
c.
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{Cu}=0,05.64=3,2g\)
Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
\(PTHH:\)
\(Fe+2HCl--->FeCl_2+H_2\left(1\right)\)
\(CuO+H_2\overset{t^o}{--->}Cu+H_2O\left(2\right)\)
a. Theo PT(1): \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,15.22,4=3,36\left(lít\right)\)
b. Ta thấy: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\)
Vậy CuO dư.
Theo PT(2): \(n_{Cu}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)
\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{Fe_2O_3} = \dfrac{3,2}{160} = 0,02(mol)\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ 3n_{Fe_2O_3} = 0,02.3 = 0,06 < n_{H_2} = 0,1 \to H_2\ dư\)
Vậy lượng sắt III oxit trên phản ứng hết với lượng hidro sinh ra.
a) PTPƯ: Zn + 2 HCl → Zn\(_{ }Cl_2\) + \(_{_{ }}H_2\)
\(_{ }n_{Zn}\) = \(\dfrac{6,5}{65}\) = 0,1 ( mol)
Theo PTPƯ: để có 1 mol \(_{_{ }}H_2\) cần 1 mol Zn
⇒ có 0,1 mol Zn sẽ tạo ra 0,1 mol \(_{_{ }}H_2\)
\(_{ }V_{H_2}\) = n. 22,4 = 0,1 . 22,4 = 2,24 ( l)
c)
PTPƯ: 3 \(_{ }H_2\) + \(_{ }Fe_2O_3\) → 3 \(_{ }H_2O\) + 2Fe
tỉ lệ: 3 : 1 : 3 : 2
Số mol: 0,1 : \(\dfrac{1}{30}\)
\(_{ }m_{Fe_2O_3}\) = \(\dfrac{1}{30}\) . 160 = 5,3 ( g)
nZn = 52 : 65 = 0,8 (mol)
pthh : Zn + 2HCl ---> ZnCl2 +H2
0,8--->1,6----------------->0,8 (mol)
=> mHCl = 1,6 . 36,5 = 58,4 (g)
VH2 = 0,8 . 22,4 = 17,92 (l)
nFe3O4 = 9,28 : 232 = 0,04 (mol )
pthh : Fe3O4 + 4H2 -t--> 3Fe + 4H2O
LTL :
0,04/1 < 0,8/4 => H2 DU
theo pthh , nFe = 3nFe3O4 = 0,12 (mol)
=> m Fe = 0,12 . 56= 6,72 (g)
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1\cdot22,4=2,24\left(l\right)\\ b,PTHH:CuO+H_2\rightarrow^{t^o}Cu+H_2O\\ \Rightarrow n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
`a)`
PTHH : `Zn + 2HCl -> ZnCl_2 + H_2`
`200ml = 0,2l`
`n_{HCl} = 0,2 . 1 = 0,2` `mol`
`n_{H_2} = 1/2 . n_{HCl} = 0,1` `mol`
`V_{H_2} = 0,1 . 22,4 = 2,24` `l`
`b)`
`CuO + H_2 -> Cu + H_2O`
Ta có : `n_{H_2} = 0,1` `mol`
`-> n_{Cu} = n_{H_2} = 0,1` `mol`
`-> m_{Cu} = 0,1 . 64 = 6,4` `gam`
