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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
bđ 0,3 0,4
pư 0,3 0,15
sau pư 0 0,25 0,3
=> H2 hết, O2 dư
\(m_{O_2\left(dư\right)}=0,25.32=8\left(g\right)\)
b) \(A_{H_2O}=0,3.6.10^{23}=1,8.10^{23}\left(phân.tử\right)\)
c) \(m_{O_2\left(pư\right)}=0,15.32=4,8\left(g\right)\)
PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,3<-------------------------------------0,15
\(\rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Số mol khí H2 và khí O2 lần lượt là 6,72:22,4=0,3 (mol) và 8,96:22,4=0,4 (mol).
2H2 (0,3 mol) + O2 (0,15 mol) \(\rightarrow\) 2H2O (0,3 mol). Do 0,3:2<0,4 nên sau phản ứng, khí H2 hết, khí O2 dư và dư (0,4-0,15).32=8 (g).
b) Số phân tử nước tạo ra sau phản ứng là 0,3.NA (phân tử) với NA là hằng số Avogadro.
c) 2KMnO4 (0,3 mol) \(\underrightarrow{t^o}\) K2MnO4 + MnO2 + O2 (0,15 mol).
Khối lượng cần tìm là 0,3.158=47,4 (g).
![](https://rs.olm.vn/images/avt/0.png?1311)
`4P + 5O_2` $\xrightarrow{t^o}$ `2P_2 O_5`
`0,32` `0,4` `0,16` `(mol)`
`a)n_P=[12,4]/31=0,4(mol)`
`n_[O_2]=[8,96]/[22,4]=0,4(mol)`
Có: `[0,4]/4 > [0,4]/5`
`=>O_2` hết, `P` dư
`=>m_[P(dư)]=(0,4-0,32).31=2,48(g)`
`b)m_[P_2 O_5]=0,16.142=22,72(g)`
`c)`
`2KMnO_4` $\xrightarrow{t^o}$ `K_2 MnO_4 +MnO_2 + O_2`
`0,8` `0,4` `(mol)`
`=>m_[KMnO_4]=0,8.158=126,4(g)`
`d)`
`KClO_3` $\xrightarrow[MnO_2]{t^o}$ `KCl + 3/2 O_2`
`4/15` `0,4` `(mol)`
`=>m_[KClO_3]=4/15 . 122,5~~32,67(g)`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\a, PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,2}{4}< \dfrac{0,2}{3}\Rightarrow O_2dư\\ \Rightarrow n_{O_2\left(dư\right)}=0,2-\dfrac{3}{4}.0,2=0,05\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\\ b,n_{Al_2O_3}=\dfrac{n_{Al}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,1=10,2\left(g\right)\)
a, Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,2}{3}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
b, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2.......0.4.......................0.2\)
\(m_{Zn}=0.2\cdot65=13\left(g\right)\)
\(C\%_{HCl}=\dfrac{0.4\cdot36.5}{200}\cdot100\%=7.3\%\)
\(n_{CuO}=\dfrac{24}{80}=0.3\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(1..........1\)
\(0.3.........0.2\)
\(LTL:\dfrac{0.3}{1}>\dfrac{0.2}{1}\Rightarrow CuOdư\)
\(m_{CuO\left(dư\right)}=\left(0.3-0.2\right)\cdot64=6.4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Theo gt ta có: $n_{H_2}=0,15(mol);n_{O_2}=0,05(mol)$
$2H_2+O_2\rightarrow 2H_2O$
Sau phản ứng $H_2$ còn dư. Và dư 0,05.22,4=1,12(l)
b, Ta có: $n_{H_2O}=2.n_{O_2}=0,1(mol)\Rightarrow m_{H_2O}=1,8(g)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Ta có: \(n_{H_2}=n_{O_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,25}{1}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,125\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,125=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=0,125.24,79=3,09875\left(l\right)\)
b, Theo PT: \(n_{H_2O}=n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,25.18=4,5\left(g\right)\)
c, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{6}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{6}.122,5\approx20,42\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\), ta được Fe dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\)
c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
trc p/u : 0,1 0,1
p/u : 0,05 0,1 0,05 0,05
sau p/u: 0,05 0 0,05 0,05
---> sau p/ư : Fe dư
\(a,V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, \(m_{Fedư}=0,05.56=2,8\left(g\right)\)
\(c,_{FeCl_2}=0,05.127=6,35\left(g\right)\)
\(m_{ddFeCl_2}=5,6+\left(0,1.36,5\right)-\left(0,05.1\right)=9,2\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{6,35}{9,2}.100\%\approx69\%\)
a)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\); \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,4}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,3->0,15------->0,3
=> \(m_{O_2\left(dư\right)}=\left(0,4-0,15\right).32=8\left(g\right)\)
b) Số phân tử H2O = 0,3.6.1023 = 1,8.1023 (phân tử)
c)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,3<--------------------------------0,15
=> \(m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
tớ cảm ơn nhaaa