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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_K=\dfrac{3,8}{39}=\dfrac{19}{195}mol\)
\(n_{H_2O}=\dfrac{101,8}{18}=\dfrac{509}{90}mol\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
19/195 < 509/90 ( mol )
19/195 19/195 19/195 ( mol )
Chất dư là H2O
\(m_{H_2O\left(dư\right)}=\left(\dfrac{509}{90}-\dfrac{19}{195}\right).18\approx100,04g\)
\(m_{KOH}=\dfrac{19}{195}.56\approx5,45g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(PTHH:Fe+S\underrightarrow{t^o}FeS\)
bđ______0,2___0,1
pư______0,1___0,1__0,1
kt_______0,1__0____0,1
C/m sau pư: Fe dư, S hết
\(m_{Fedư}=0,1.56=5,6g\)
\(m_{FeS}=0,1.88=8,8g\).
nFe=mFeMFe=16,856=0,3(mol)
nFe3O4=mFe3O4MFe3O4=16232=0,07(mol)
PTHH:
(mol) 1..............1...............1
(mol) 0,1........0,1...........0,1
*Tỉ lệ mol:
nFe:nFe3O4=0,33>0,071
⇒Fe dư
b) mFeS = nFeS . MFeS = 0,1 . 88 = 8,8 (g)
nFe dư = 0,2 - 0,1 = 0,1 (mol)
mFe dư = nFe dư . MFe dư = 0,1 . 56 = 5,6 (g).
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(n_{Mg}=\dfrac{35,6}{24}=1,483\left(mol\right)\)
\(V_{O2\left(đktc\right)}=\dfrac{21,504}{22,4}=0,96\left(mol\right)\)
pt: 2Mg + O2 → 2MgO (1)
mol: 2 1 2
mol:1,483 0,96
Tỉ lệ: \(\dfrac{1,483}{2}=0,7415< \dfrac{0,96}{1}=0,96\)
Mg tác dụng hết. O2 dư
theo PTHH có
\(n_{O2p\intư}=\dfrac{1,843x1}{2}=0,7415\left(mol\right)\)
nO2 dư=1,843-0,7415=1,1015 (mol)
mO2dư= 1,1015 x 32 = 35,48 (g)
b)theo PTHH có
\(n_{MgO}=\dfrac{1,843x2}{2}=1,843\left(mol\right)\)
nMgO = 1,843 X 40 = 73,72 (g)
c)
nMg PT(1)=nMgPT(2)=1,843 (mol)
pt: Mg + H2SO4 ➝ MgSO4 + H2 (2)
mol: 1 1 1 1
mol: 1,843
Theo PTHH có
\(n_{H2}=\dfrac{1,843x1}{1}=1,843\) (mol)
mH2=1,843 x 2 = 3,686 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. PTHH: \(2H_2+O_2\rightarrow2H_2O\)
b. \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)
Theo PTHH: \(n_{H_2}=n_{O_2}\cdot\dfrac{2}{1}=0,1\cdot\dfrac{2}{1}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c. Chưa hiểu đề bài lắm :))
d. Ta có: \(n_{O_2}=0,1\left(mol\right);n_{H_2}=\dfrac{3,6}{22,4}\approx0,16\left(mol\right)\)
Do \(0,1< 0,16\) nên \(H_2\) dư \(0,16-0,1=0,06\left(mol\right)\)
Câu c dễ hiểu mà em, không phải sản phẩm tạo nước à em
![](https://rs.olm.vn/images/avt/0.png?1311)
\(nAl=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(nHCl=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2 6 2 3 (mol)
0,2 0,6 0,2 0,3 (mol)
LTL : 0,3 / 2 > 0,6/6
=> Al dư sau pứ , HCl đủ vs pứ
\(mAl_{\left(dư\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
\(mAlCl_3=0,2.98=19,6\left(g\right)\)
\(H_2+CuO\rightarrow Cu+H_2O\)
1 1 1 1 (mol)
0,3 0,3 0,3 0,3 (mol)
=> \(mCu=0,3.64=19,2\left(g\right)\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\
n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\
pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(LTL:\dfrac{0,3}{2}>\dfrac{0,6}{6}\)
=> Al dư HCl hết
theo pthh : \(n_{Al\left(p\text{ư}\right)}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\\ m_{Al\left(d\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
theo pthh : \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,1\left(mol\right)\\
m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
theo pthh : \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
pthh: \(CuO+H_2\underrightarrow{t^o}H_2O+Cu\)
0,3 0,3
\(m_{Cu}=0,3.64=19,2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. PTHH: 2Al(OH)3 + 3H2SO4 ---> Al2(SO4)3 + 6H2O
b. Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
Ta thấy: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\)
Vậy \(Al\left(OH\right)_3\) dư.
\(m_{dư}=0,75.78-98.0,5=9,5\left(g\right)\)
c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,5=\dfrac{1}{6}\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right);n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Mol: \(\dfrac{1}{3}\) 0,5 \(\dfrac{1}{6}\)
b, Ta có: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\) ⇒ Al(OH)3 dư, H2SO4 hết
⇒ \(m_{Al\left(OH\right)_3}=\left(0,75-\dfrac{1}{3}\right).78=32,5\left(g\right)\)
c, \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)
Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)
a. PTHH: 2Al(OH)3 + 3H2SO4 ---> Al2(SO4)3 + 6H2O
b. Không có chất dư (hoặc có thể bn cho sai 49(g) dung dịch là 49(g) H2SO4)
c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,75=0,375\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,375.342=128,25\left(g\right)\)
theo bài ra ta có :
mS = 3,2g => nS = 0,1 mol
mFe = 11,2 g => nFe = 0,2mol
pthh:
Fe + S - > FeS
1mol......1mol.......1mol
0,2mol....0,1mol
theo pt: nFe = nS
theo gt: nFe>nS ( 0,2>0,1)
=> Fe dư
theo pthh ta có nFeS = nS = 0,1mol
=> mFeS = 0,1 . 88 = 8,8(g)
ta có nFe(dư) = 0,2 - 0,1 = 0,1 mol
=> mFe(dư) = 0,1 . 56 = 5,6 (g)