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a)\(Fe2O3+3H2-->2Fe+3H2O\)
b)\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{Fe}=\frac{2}{3}n_{H2}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
c)\(Fe+H2SO4-->FeSO4+H2\)
\(n_{H2SO4}=n_{Fe}=0,2\left(mol\right)\)
\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,35}{1}\), ta được H2 dư.
Theo PT: \(n_{H_2\left(pư\right)}=n_{CuO}=0,15\left(mol\right)\)
\(\Rightarrow n_{H_2\left(dư\right)}=0,35-0,15=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2\left(dư\right)}=0,2.2=0,4\left(g\right)\)
a, PT: Mg + 2HCl ---> MgCl2 + H2
b, Số mol của Mg là:
n = \(\frac{m}{M}\)= \(\frac{4,8}{24}\)= 0,2 ( mol )
Theo PT, ta có: nH2 = nMg = 0,2 (mol )
Thể tích khí sinh ra là:
V = n. 22,4= 0,2. 22,4= 4,48 ( l )
c, PT: 3H2 + Fe2O3 --to--> 2Fe + 3H2O
Số mol của Fe2O3 là:
n= \(\frac{m}{M}\)= \(\frac{16}{160}\)= 0,1 ( mol )
nH2 : nFe2O3 = \(\frac{0,2}{3}\): \(\frac{0,1}{1}\)= 0,067 < 0,1
Vậy Fe2O3 dư, tính theo H2
Theo PT, ta có:
nFe = nH2 = 0,2 ( mol )
Khối lượng sắt tạo thành là:
m = n. M= 0,2. 56 = 11,2 (g )
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\left(1\right)\)
b) \(n_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\)
Theo PTHH: \(n_{H_2}:n_{Mg}=1:1\)
\(\Rightarrow n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{Fe_2O_3}=\frac{16}{160}=0,1\left(mol\right)\)
\(n_{H_2}=0,2\left(mol\right)\)
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^0}2Fe+3H_2O\left(2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{n_{Fe_2O_3}}{1}=\frac{0,1}{1}=0,1\\\frac{n_{H_2}}{3}=\frac{0,2}{3}=0,07\end{matrix}\right.\) \(\Rightarrow Fe_2O_3\) dư. H2 phản ứng hết vậy tính toán theo \(n_{H_2}\)
Theo PTHH: \(n_{H_2}:n_{Fe}=3:2\)
\(\Rightarrow n_{Fe}=n_{H_2}.\frac{2}{3}=0,2.\frac{2}{3}=0,13\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,13.56=7,28\left(g\right)\)
a, \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
PTHH: Mg + H2SO4 ---> MgSO4 + H2
0,3--->0,3--------------------->0,3
=> mH2SO4 = 0,3.98 = 29,4 (g)
b, VH2 = 0,3.22,4 = 6,72 (l)
c, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
LTL: \(0,2>\dfrac{0,3}{3}\) => Fe2O3 dư
Theo pthh: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
=> mFe2O3 (dư) = (0,2 - 0,1).160 = 16 (g)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,3 0,3 0,3
\(m_{H_2SO_4}=0,3.98=29,4g\\
V_{H_2}=0,3.22,4=6,72l\)
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\
pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(LTL:\dfrac{0,2}{1}>\dfrac{0,3}{3}\)
=> Fe dư
\(n_{Fe\left(p\text{ư}\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\\
m_{Fe\left(d\right)}=\left(0,2-0,1\right).56=5,6g\)
a) \(Fe+H2SO4-->FeSO4+H2\)
b)\(nFe=\frac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H2}=n_{Fe}=0,2\left(mol\right)\)
\(VH2=0,2.22,4=4,48\left(l\right)\)
c)\(n_{FeSO4}=n_{Fe}=0,2\left(mol\right)\)
\(m_{FeSO4}=0,2.152=30,4\left(g\right)\)
d)\(CuO+H2-->Cu+H2O\)
\(n_{Cu}=n_{H2}=0,2\left(mol\right)\)
\(m_{Cu}=0,2.64=12,8\left(g\right)\)
\(n_{Mg}=\frac{m}{M}=\frac{9,6}{24}=0,4mol\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
1 : 1 : 1 : 1 mol
0,4 0,4 0,4 0,4 mol
a. \(m_{MgSO_4}=n.M=0,4.\left(24+32+16.4\right)=48g\)
b. \(V_{H_2}=n.22,4=0,4.22,4=8,96l\)
c. \(n_{Fe_2O_3}=\frac{m}{M}=\frac{64}{56.2}+16.3=0,4mol\)
PTHH: \(3H_2+Fe_{2O_3}\rightarrow2Fe+3H_2O\left(ĐK:t^o\right)\)
3 : 1 : 2 : 3 mol
1, 7 0,4 0,8 1,2 mol
\(m_{Fe}=n.M=0,8.56=44,8g\)
a)\(Fe3O4+4H2-->3Fe+4H2O\)
b)\(n_{H2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
\(n_{Fe3O4}=\frac{1}{4}n_{H2}=0,1\left(mol\right)\)
\(m_{Fe3O4}=0,1.232=23,2\left(g\right)\)
c) \(Fe+H2SO4-->FeSO4+H2\)
\(n_{Fe}=\frac{3}{4}n_{H2}=0,3\left(mol\right)\)
\(n_{H2}=n_{Fe}=0,3\left(mol\right)\)
\(V_{H2}=0,3.22,4=6,72\left(l\right)\)
Bài làm
2KClO3 -----> 2KCl + 3O2
a) nKClO3 = 19,6/( 39 + 35,5 + 16 . 3 ) = 0.16 ( mol )
nO2 = 3/2 nKClO3 = 3/2 . 0,16 = 0,24 mol
VO2 = 0,24 . 22,4 = 5,376 ( l )
b) 2O2 + 3Fe ---> Fe3O4
nFe3O4 = 1/2 nO2 = 1/2 . 0,24 = 0,12 mol
=> mFe3O4 = 0,12 . ( 56 . 3 + 16 . 4 ) = 27,84 ( g )
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)
a)\(Mg+H2SO4-->MgSO4+H2\)
b)\(n_{Mg}=\frac{12}{24}=0,5\left(mol\right)\)
\(n_{H2}=n_{Mg}=0,5\left(mol\right)\)
\(V_{H2}=0,5.22,4=11,2\left(l\right)\)
c)\(4H2+Fe3O4-->3Fe+4H2O\)
\(n_{Fe}=\frac{3}{4}n_{H2}=0,375\left(mol\right)\)
\(m_{Fe}=0,375.56=21\left(g\right)\)
a) nMg= 12/24= 0,5(mol)
PTHH: Mg + H2SO4 -> MgSO4 + H2
0,5________0,5_______0,5_____0,5(mol)
b) V(H2,đktc)= 0,5.22,4= 11,2(l)
c) 4 H2 + Fe3O4 -to-> 4 H2O + 3 Fe
0,5_______________________0,375
=> mFe= 0,375. 56= 21(g)