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\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2.......0.4.......................0.2\)
\(m_{Zn}=0.2\cdot65=13\left(g\right)\)
\(C\%_{HCl}=\dfrac{0.4\cdot36.5}{200}\cdot100\%=7.3\%\)
\(n_{CuO}=\dfrac{24}{80}=0.3\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(1..........1\)
\(0.3.........0.2\)
\(LTL:\dfrac{0.3}{1}>\dfrac{0.2}{1}\Rightarrow CuOdư\)
\(m_{CuO\left(dư\right)}=\left(0.3-0.2\right)\cdot64=6.4\left(g\right)\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right);n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\\ a,Vì:\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCldư\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\\ n_{HCl\left(dư\right)}=0,8-0,3.2=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{MgCl_2}=0,3.95=28,5\left(g\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
a) \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
\(n_{HCl}=\dfrac{29.2}{36,5}=0,8\left(mol\right)\)
PTHH : 2Mg + 2HCl -> 2MgCl + H2
Xét tỉ lệ \(\dfrac{0,3}{2}< \dfrac{0,8}{2}\)
=> HCl dư
=> \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
=> \(V_{MgCl}=0,15.22,4=3,36\left(l\right)\)
b) \(m_{H_2}=0,075.2=0,15\left(g\right)\\ m_{MgCl}=0,15.59,5=8,925\left(g\right)\)
a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4------>0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mHCl(PTHH) = 0,4.36,5 = 14,6 (g)
=> \(m_{HCl\left(tt\right)}=\dfrac{14,6.120}{100}=17,52\left(g\right)\)
c)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,2--->0,1------->0,2
=> mH2O = 0,2.18 = 3,6 (g)
mO2(dư) = (0,2 - 0,1).32 = 3,2(g)
nZn = 13/65 = 0,2 (mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2
VH2 = 0,2 . 22,4 = 4,48 (l)
mHCl = (0,4 . 36,5)/(100% + 20%) = 73/6 (g)
nO2 = 4,48/22,4 = 0,2 (mol)
PTHH: 2H2 + O2 -> (t°) 2H2O
LTL: 0,2/2 < 0,2 => O2 dư
nH2O = nH2 = 0,2 (mol)
mH2O = 0,2 . 18 = 3,6 (g)
- pt: Zn + 2HCl -> ZnCl2 +H2
- nHCl = ( 3,25 : 65 ) x 2 = 0,1 (mol)
V = 0,1 : 0,5 = 0,2 (l)
- gọi a là số mol cần tìm
- pt: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
a -> 3/2a
Fe + H2SO4 -> FeSO4 + H2
a -> a
- ta có : a + 3/2a = 0,05 => a = 0,02 (mol)
- C%Fe = ( 0,02 x 56)x100 / (0,02x56 + 0,02x 27) = 67,47%
- C% Al = 100 -67,47= 32,53%
\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = \dfrac{8,1}{36,5} = \dfrac{81}{365}(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ \dfrac{n_{Al}}{2} = 0,05 > \dfrac{n_{HCl}}{6} = \dfrac{27}{730} \to Al\ dư\\ n_{Al\ pư} = \dfrac{1}{3}n_{HCl} = \dfrac{27}{365}(mol)\\ \)
\(m_{Al\ dư} = 2,7 - \dfrac{27}{265}.27 = 0,703(gam)\)
`a)PTHH:`
`Fe + 2HCl -> FeCl_2 + H_2`
`0,3` `0,6` `0,3` `0,3` `(mol)`
`n_[Fe]=[22,4]/56=0,4(mol)`
`n_[HCl]=0,3.2=0,6(mol)`
Ta có:`[0,4]/1 > [0,6]/2`
`=>Fe` dư
`b)m_[FeCl_2]=0,3.127=38,1(g)`
`c)m_[Fe(dư)]=(0,4-0,3).56=5,6(g)`
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Xét: \(\dfrac{0,4}{1}>\dfrac{0,6}{2}\) ( mol )
0,3 0,6 0,3 ( mol )
\(m_{FeCl_2}=0,3.127=38,1\left(g\right)\)
\(m_{Fe\left(dư\right)}=\left(0,4-0,3\right).56=5,6\left(g\right)\)
1/ Fe +2 HCl --------> FeCl2 + H2
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(n_{HCl}=2n_{Fe}=0,1\left(mol\right)\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)
Câu 6 :
1) $n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$
Fe + 2HCl → FeCl2 + H2
0,05...0,1....................0,05......(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$
$m_{HCl} = 0,1.36,5 = 3,65(gam)$
2)
a) $CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$V_{O_2} = 2V_{CH_4} = 4(lít)$
b) $n_{CO_2} = n_{CH_4} = 0,15(mol) \Rightarrow V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c) $d_{CH_4/kk} = \dfrac{16}{29} = 0,552$
Vậy khí metan nhẹ hơn không khí 0,552 lần
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ LTL:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(pứ\right)}=2n_{Zn}=0,4\left(mol\right)\\\Rightarrow m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\\ b.n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4,=4,48\left(l\right)\\ d.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O \\ n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\\ LTL:\dfrac{0,2}{3}< \dfrac{0,12}{1}\Rightarrow Fe_2O_3dưsauphảnứng\\ \Rightarrow n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{2}{15}.56=7,467\left(g\right)\)
a) n\(Zn\)=\(\dfrac{m}{M}\)=\(\dfrac{13}{65}\)=0,2(mol)
n\(HCl\)=\(\dfrac{m}{M}\)=\(\dfrac{18,25}{36,5}=\)0,5(mol)
PTHH : Zn + 2HCl->ZnCl\(2\) + H\(2\)
0,2 0,5
Lập tỉ lệ mol : \(^{\dfrac{0,2}{1}}\)<\(\dfrac{0,5}{2}\)
n\(Zn\) hết , n\(HCl\) dư
-->Tính theo số mol hết
Zn + 2HCl->ZnCl\(2\) + H\(2\)
0,2 -> 0,4 0,2 0,2
n\(HCl\) dư= n\(HCl\)(đề) - n\(HCl\)(pt)= 0,5 - 0,4 = 0,1(mol)
m\(HCl\) dư= 0,1.36,5 = 3,65(g)
b) m\(ZnCl2\) = n.M= 0,2.136= 27,2 (g)
c)V\(H2\)=n.22,4=0,2.22,4=4,48(l)
d) n\(Fe\)\(2\)O\(3\)=\(\dfrac{m}{M}\)=\(\dfrac{19,2}{160}\)=0,12 (mol)
3H2 +Fe2O3 → 2Fe + 3H2O
0,2 0,12
Lập tỉ lệ mol: \(\dfrac{0,2}{3}\)<\(\dfrac{0,12}{1}\)
nH2 hết .Tính theo số mol hết
\(HCl\)
3H2 +Fe2O3 → 2Fe + 3H2O
0,2-> 0,2
m\(Fe\)=n.M= 0,2.56= 11,2(g)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,5 0,25 0,25
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(a,m_{ZnCl_2}=0,25.136=34\left(g\right)\)
\(b,m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{7,3}=250\left(g\right)\)
\(c,2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
0,5 0,25
\(m_K=39.0,5=19,5\left(g\right)\)
nZn = 0,1969 mol
Zn + 2HCl -----> ZnCl2 + H2
- Theo PTHH(1): nH2 = 0,1969 mol
2H2 + O2 ---to-----> 2H2O
- Theo PTHH(2): nH2O = 0,1969 mol
=> mH2O = 0,1969 . 18 = 3,5442 gam
Bài yêu cầu tính nồng độ mol mà