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\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)
\(0.1.......0.075.....0.05\)
\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)
\(m_{Fe_2O_3}=0.05\cdot160=8\left(g\right)\)
PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)
\(n_{H_2O}=\dfrac{1.8}{18}=0.1\left(mol\right)\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)
\(0.1.........0.25......0.2..........0.1\)
\(V_{C_2H_2}=2.24\left(l\right)\)
\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)
Ta có :
\(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ n_{O_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
Ta thấy : \(\dfrac{n_{Al}}{4} = 0,025 < \dfrac{n_{O_2}}{3} =0,03\) nên O2 dư.
\(n_{Al_2O_3} = 0,5n_{Al} = 0,05(mol)\\ \Rightarrow m_{Al_2O_3} = 0,05.102 = 5,1(gam)\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(n_{O_2\left(đktc\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)
Ta có tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,1}{3}\)
-> \(O_2\) sẽ dư sau phản ứng.
Theo pthh: \(n_{Al_2O_3}=\dfrac{2}{4}n_{Al}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
-> \(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
$a\big)$
Bảo toàn KL:
$m_{O_2}=m_{CR}-m_{Fe}=11,84-11,2=0,64(g)$
$\to n_{O_2}=\frac{0,64}{32}=0,02(mol)$
$\to V_{O_2}=0,02.22,4=0,448(l)$
$b\big)$
$V_{kk}=5V_{O_2}=5.0,448=2,24(l)$
1. 2Al + 3O2 \(\rightarrow\) 2AlCl3
tỉ lệ 2:3:2
2. 2Fe + 3Cl2 \(\rightarrow\) 2FeCl3
tỉ lệ 2:3:2
3. 2Na + 2H2O \(\rightarrow\) 2NaOH + H2
tỉ lệ 2:2:2
4. Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
tỉ lệ 1:2:1:1
5. 2CxHy + (2x+y/2)O2 \(\rightarrow\) 2xCO2 + yH2O
tỉ lệ 2:(2x+y/2):2x:y
6.P2O5 + 3H2O \(\rightarrow\) 2H3PO4
tỉ lệ 1:3:2
7. Fe2(SO4)3 + 6KOH \(\rightarrow\) 2Fe(OH)3 + 3K2SO4
tỉ lệ 1:6:2:3
8. 2Fe + 3Cl2 \(\rightarrow\) 2FeCl3
tỉ lệ 2:3:2
9. 2CnH2n-2 + (n-1)O2 \(\rightarrow\) 2nCO2 + 2(n-1)H2O
tỉ lệ 2:(n-1):2n:2(n-1)
10. N2O5 + H2O \(\rightarrow\) 2HNO3
tỉ lệ 1:1:2
11. FeCl3 + 3NaOH \(\rightarrow\) Fe(OH)3 + NaCl
tỉ lệ 1:3:1:1
12. 2Al + 3Cl2 \(\rightarrow\) 2AlCl3
tỉ lệ 2:3:2
4P+5O2−>2P2O5
2Fe(OH)3−>Fe2O3+3H2O
Al2O3+6HCl−>2AlCl3+3H2O
2K+2H2O−>2KOH+H2
4Na+O2−−−>2Na2O
2KClO3−−−>2KCl+3O2
a, 4P+5O2--->2P2O5
Tỉ lệ: 4:5:2
b, 2Fe(OH)3--->Fe2O3+3H2O
Tỉ lệ: 2:1:3
c, Al2O3+6HCl--->2AlCl3+3H2O
Tỉ lệ: 1:6:2:3
d, 2K+2H2O--->2KOH+H2
Tỉ lệ: 2:2:2:1
e, 4Na+O2--->2Na2O
Tỉ lệ: 4:1:2
f, 2KClO3--->2KCl+3O2
Tỉ lệ: 2:2:3
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.1....0.075.....0.05\)
\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)
\(m_{Al_2O_3}=0.05\cdot102=5.1\left(g\right)\)
ta có: \(n_{Al}=0.1\left(mol\right)\)
PTHH
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
4 3
0.1 x
\(=>x=0.075=n_{O_2}\)
\(=>V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)