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`3Fe+2O_2 \overset{t^o}\to Fe_3O_4`
`n_{O_2}=0,2(mol)`
`=>n_{Fe_3O_4}=1/2n_{O_2}=0,1(mol)`
`=>m_{Fe_3O_4}=23,2`
`n_{Fe}=3/2n_{O_2}=0,3(mol)`
`=>n_{Fe}=16,8`
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a) Ta có: \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow n_{Fe}=0,45mol\) \(\Rightarrow m_{Fe}=0,45\cdot56=25,2\left(g\right)\)
b)
+) Cách 1:
Theo PTHH: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,15mol\)
\(\Rightarrow m_{Fe_3O_4}=0,15\cdot232=34,8\left(g\right)\)
+) Cách 2:
Ta có: \(m_{O_2}=0,3\cdot32=9,6\left(g\right)\)
Bảo toàn khối lượng: \(m_{Fe_3O_4}=m_{O_2}+m_{Fe}=25,2+9,6=34,8\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{4,958}{22,4}\approx0,2\left(mol\right)\)
\(PTHH:3Fe+2O_2-^{t^o}>Fe_3O_4\)
tỉ lệ 3 : 2 : 1
BĐ 0,5 0,2
PU 0,3----->0,2--------->0,1
CL 0,2------>0----------->0,1
có
\(\dfrac{n_{Fe}}{3}>\dfrac{n_{O_2}}{2}\left(\dfrac{0,5}{3}>\dfrac{0,2}{2}\right)\)
Fe dư, `O_2` hết, tính theo`O_2`
\(m_{Fe_3O_4}=n\cdot M=0,1\cdot\left(56\cdot3+16\cdot4\right)=23,2\left(g\right)\)
nFe = 2.8/56 = 0.05 (mol)
nO2 = 22.4 / 22.4 = 1 (mol)
3Fe + 2O2 -to-> Fe3O4
0.05__1/30______1/60
mO2 (dư) = ( 1 - 1/30) * 32 = 30.93 (g)
mFe3O4 = 1/60 * 232 = 3.867 (g)
a/ \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b/ Ta có: \(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)
\(n_{O_2}=\dfrac{22.4}{22.4}=1\left(mol\right)\)
Ta có: \(\dfrac{n_{Fe\left(bra\right)}}{n_{Fe\left(pt\right)}}=\dfrac{0.05}{3}=0.016< \dfrac{n_{O_2\left(bra\right)}}{n_{O_2\left(pt\right)}}=\dfrac{1}{2}=0.5\)
=> Oxi phản ứng dư
mO2 dư = (1 - 1/30) . 32 = 30.93 (g)
mFe3O4 = 1/60 . 232 = 3.867 (g)
2 KClO 3 → t o 2 KCl + 3 O 2 2 O 2 + 3 Fe → t o Fe 3 O 4 Fe 3 O 4 + 4 H 2 → 3 Fe + 4 H 2 O Fe + 2 HCl → FeCl 2 + H 2 ↑
\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
tỉ lệ 2 : 2 : 3
n(mol) `1/3`<------------`1/3`<-----`0,5`
\(m_{KClO_3}=n\cdot M=\dfrac{1}{3}\cdot\left(39+35,5+16\cdot3\right)\approx40,83\left(g\right)\)
Cách 1 :
PTHH : \(3Fe+2O_2\rightarrow Fe_3O_4\)
..............0,3........0,2........0,1..........
\(n_{O_2}=\frac{V}{22,4}=0,2\left(mol\right)\)
=> \(\left\{{}\begin{matrix}m_{Fe}=n.M=16,8\left(g\right)\\m_{Fe_3O_4}=n.M=23,2\left(g\right)\end{matrix}\right.\)
Cách hai :
\(n_{O_2}=\frac{V}{22,4}=0,2\left(mol\right)\)
=> \(m_{O_2}=n.M=6,4\left(g\right)\)
-> \(n_{\left(O\right)}=0,4\left(mol\right)\)
=> \(n_{Fe_3O_4}=\frac{1}{4}n_{\left(O\right)}=0,1\left(mol\right)\)
=> \(m_{Fe_3O_4}=n.M=23,2\left(g\right)\)
- Định luật bảo toàn khối lượng :
\(m_{Fe}+m_{O_2}=m_{Fe_3O_4}\)
=> mFe = 16,8 ( g )
tại sao n(O)=0.4 (mol) v bn