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nFe = 33,6 : 56 = 0,6 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,6--> 0,4------->0,2 (mol)
=> vO2 = 0,4.22,4 = 8,96 (mol)
=> mFe3O4 = 0,2.232 = 46,4 (g)
pthh : 2KClO3 -t--> 2KClO3 + 3O2
0,267<-----------------------0,4(mol)
mKClO3= 0,267 .122,5 = 32,67 (g)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,3\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KClO_3}=0,4.122,5=49\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
\(a.PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 2 1
0,9 0,6 0,3
\(b.V_{O_2}=n.24,79=0,6.24,79=14,874\left(l\right)\)
\(c.m_{Fe_3O_4}=n.M=0,3.\left(56.3+16.4\right)=69,6\left(g\right)\)
\(d.V_{O_2}=14,874\left(l\right)\\ \Rightarrow n_{O_2}=\dfrac{V}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2 2 3
0,6 0,6 0,9
\(m_{KClO_3}=n.M=0,6.\left(39+35,5+16.3\right)=55,5\left(g\right).\)
a. \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH : 3Fe + 2O2 -to> Fe3O4
0,3 0,2 0,1
b. \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
c. \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
a \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b \(\Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3mol\) \(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1mol\Rightarrow m_{Fe_3O_4}=0,1\cdot232=2,32g\)
c \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)
\(a/3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ b/n_{Fe}=\dfrac{1,4}{56}=0,025mol\\ n_{O_2}=\dfrac{0,025.2}{3}=\dfrac{0,05}{3}mol\\ V_{O_2}=\dfrac{0,05}{3}\cdot22,4\approx0,37l\\ c/C_1\\ n_{Fe_3O_4}=\dfrac{0,025}{3}mol\\ m_{Fe_3O_4}=\dfrac{0,025}{3}\cdot232\approx1,93g\\ C_2\\ m_{O_2}=\dfrac{0,05}{3}\cdot32\approx0,53g\\ BTKL:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{Fe_3O_4}=1,4+0,53=1,93g\)
nFe = 16,8 : 56 = 0,3 (mol)
pthh :3 Fe + 2O2 -t--> Fe3O4
0,3--------------> 0,1 (mol)
=> mFe3O4 =0,1 . 232 = 23,2(G)
nH2 = 44,8 : 22,4 = 2 (g)
pthh : Fe3O4 + H2 -t--> Fe + H2O
LTL : 0,1 / 1 < 2 /1
=> H2 du
nH2 (pu) = nFe3O4 = 0,1 (mol)
=> nH2 (d) = 2-0,1 = 1,9 (mol)
mH2 (d) = 1,9 . 2 = 3,8 (g)
nFe3O4 = 23.2/232 = 0.1 mol
3Fe + 2O2 -to-> Fe3O4
0.3____0.2_______0.1
mFe = 0.3*56 = 16.8 g
VO2 = 0.2*22.4 = 4.48 (l)
Vkk = 5VO2 = 22.4 (l)
\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,3 0,2 0,1 ( mol )
\(\left\{{}\begin{matrix}m_{O_2}=0,2.32=6,4g\\V_{O_2}=0,2.22,4=4,48l\end{matrix}\right.\)
\(m_{Fe_3O_4}=0,1.232=23,2g\)
REFER
a)3 Fe+2O2--->Fe3O4
b) Ta có
n Fe=16,8/56=0,3(mol)
Theo pthh
n O2=2/3n Fe=0,2(mol)
V O2=0,2.22,4=4,48(l)
c) Cách 1
Áp dụng định luật bảo toàn khối lượng ta có
m Fe3O4=m Fe+m O2
=16,8+0,2.32=23,2(g)
Cách 2
Theo pthh
n Fe3O4=1/3n Fe=0,1(mol)
m Fe3O4=0,1.232=23,2(g)
nFe3O4= 0,1(mol)
PTHH: 3 Fe +2 O2 -to-> Fe3O4
nFe=3.nFe3O4=3.0,1=0,3(mol)
=> mFe=0,3.56=16,8(g)
nO2=2.nFe3O4=2.0,1=0,2(mol)
=>V(O2,đktc)=0,2.22,4=4,48(l)
Bài 3:
Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
_____0,2____0,6____0,4 (mol)
\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)
\(m_{Fe}=0,4.56=22,4\left(g\right)\)
Bài 4:
a, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,35}{5}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,35-0,25=0,1\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,1.32=3,2\left(g\right)\)
b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
Lần sau bạn nên chia nhỏ câu hỏi ra nhé.
Bài 1:
a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KMnO_4}=0,4.158=63,2\left(g\right)\)
Bài 2:
a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_____0,1___________0,1_____0,15 (mol)
\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{CuO}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)
PTHH:\(3Fe+2O_2\rightarrow Fe_3O_4\)
số mol Fe là:\(n=\frac{16.8}{56}=0.3\left(mol\right)\)
\(m_{Fe_3O_4}=0.1\cdot\left(56\cdot3+16\cdot4\right)=23.2\left(g\right)\)
\(V_{O_2}=\frac{\left(0.3\cdot2\right)}{3}\cdot22.4=4.48\left(l\right)\)