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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)
1,3<---4<-------1,3<---------2
b)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)
\(m_{AlCl_3}=n\cdot M=1,3\cdot\left(27+35,5\cdot3\right)=173,55\left(g\right)\)
\(m_{Al}=n\cdot M=1,3\cdot27=35,1\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.n_{H_2}=n_{Fe}=\dfrac{10}{56}=\dfrac{5}{28}\left(mol\right)\\ \Rightarrow V_{H_2}=\dfrac{5}{28}.22,4=4\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
224ml = 0,224l
\(n_{H2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)
Pt : \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2|\)
1 2 1 1
0,01 0,01 0,01
\(BaO+H_2O\rightarrow Ba\left(OH\right)_2|\)
1 1 1
0,01 0,01
\(n_{Ba}=\dfrac{0,01.1}{1}=0,01\left(mol\right)\)
\(m_{Ba}=0,01.137=1,37\left(g\right)\)
\(m_{BaO}=2,9-1,37=1,53\left(g\right)\)
0/0Ba = \(\dfrac{1,37.100}{2,9}=47,24\)0/0
0/0BaO = \(\dfrac{1,53.100}{2,9}=52,76\)0/0
Có : \(m_{BaO}=1,53\left(g\right)\)
\(n_{BaO}=\dfrac{1,53}{153}=0,01\left(mol\right)\)
\(n_{Ba\left(OH\right)2\left(tổng\right)}=0,01+0,01=0,02\left(mol\right)\)
⇒ \(m_{Ba\left(OH\right)2}=0,02.171=3,42\left(g\right)\)
Chúc bạn học tốt
Bài 2:
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,1 0,2 0,1 0,1
Ta có: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) ⇒ Zn pứ hết, HCl dư
\(m_{HCldư}=\left(0,4-0,2\right).36,5=7,3\left(g\right)\)
b, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, \(m_{dd.sau.pứ}=6,5+200-0,1.2=206,3\left(g\right)\)
\(C\%_{HCldư}=\dfrac{7,3.100\%}{206,3}=3,54\%\)
\(C\%_{ZnCl_2}=\dfrac{0,1.136.100\%}{206,3}=6,59\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)PTHH:`
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`0,2` `0,6` `0,2` `0,3` `(mol)`
`n_[Al] = [ 5,4 ] / 27 = 0,2 (mol)`
`b) V_[H_2] = 0,3 . 22,4 = 6,72 (l)`
`c) V_[dd HCl] = [ 0,6 ] / [ 2,4 ] = 0,25 (l)`
`=> C_[M_[AlCl_3]] = [ 0,2 ] / [ 0,25 ] = 0,8 (M)`
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,2 0,3 ( mol )
\(V_{H_2}=0,3.22,4=6,72l\)
\(V_{HCl}=\dfrac{0,6}{2,4}=0,25l\)
\(C_{M_{AlCl_3}}=\dfrac{0,2}{0,25}=0,8M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Fe+2HCl--->FeCl2+H2
n Fe=28/56=0,5(mol)
Theo pthh
n H2=n Fe=0,5(mol)
V H2=0,5.22,4=11.2(l)
b) 3H2+Fe2O3---->2Fe+3H2O
n Fe=48/160=0,3(mol)
n H2=0,5(mol)
---->Fe dư
Theo pthh
n Fe=2/3n H2=0,333(mol)
m Fe=0,3333.56=18,67(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
CuO +H2 --> Cu +H2O (1)
Fe2O3 +3H2 --> 2Fe + 3H2O (2)
nH2=19,6/22,4=0,875(mol)
mH2=0,875.2=1,75(g)
giả sử nCuO=x(mol)
nFe2O3=y(mol)
=>80x +160y=50(I)
theo (1) : nH2=nCuO=x(mol)
theo(2) : nH2=3nFe2O3=3y(mol)
=> 2x+6y=0,875 (II)
từ (I) và (II) ta có :
80x +160y=50
2x +6y=0,875
hình như sai đề
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : nH2 = VH2 / 22,4 = 1,12 / 22,4 = 0,05 mol ( ĐKTC )
Theo PTHH : nR = nH2 = 0,05 ( mol )
Mà nR = mR / MR = 4,8 / MR ( mol )
-> 4,8 / MR = 0,05
-> MR = 96
-> R là Cm
Chắc gì R là 1 nguyên tố hóa học,lỡ nó là hợp chất thì sao ?
\(Ba+2HCl\left(0,1\right)\rightarrow BaCl_2\left(0,05\right)+H_2\left(0,05\right)\)
\(m_{HCl}=50.7,3\%=3,65g\Rightarrow n_{HCl}=0,1mol\)
\(\Rightarrow V_{H_2}=0,05.22,4=1,12l\)
\(m_{BaCl_2}=0,05.208=10,4g\)
a)mHCl = 50*7.3/100=3.65g
nHCl = 3.65/36.5=0.1 mol
Ba + 2HCl -> BaCl2 + H2
(mol) 0.1 0.05 0.05
vH2= 0.05*22.4 = 1.12(l)
b) mBaCl2 = 0.05*208=10.4g