cho 16,8g Fe cháy trong 7,437 l O2, sau phản ứng Fe hay O2 dư? tính lượng dư? tính msp (Fe3O4)
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PTHH: \(4Fe+3O_2\rightarrow2Fe_2O_3\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,3}{4}>\dfrac{0,1}{2}\) \(\Rightarrow\) Sắt còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{2}n_{Fe_2O_3}=0,15\left(mol\right)\\n_{Fe\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(dư\right)}=0,1\cdot56=5,6\left(g\right)\\V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)
\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(3.........2\)
\(0.225......0.1875\)
Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)
\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)
\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)
Ta có: nFe = 22,4/56 = 0,4 (mol)
nO2 = 4,48/22,4 = 0,2 (mol)
PTHH: 3Fe + 2O2 -> Fe3O4
3 mol 2mol 1mol
SS: 0,4/3mol > 0,2/2mol -> Fe dư , O2 hết
Ta có: nFe p/ứng là: 0,2.3/2 = 0,3 mol
=> mFe phản ứng: 0,3.56 =16,8 (g)
=> mFe dư: 22,4 - 16,8 = 5,6 (g)
\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)
\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)
\(0.2....0.15.........0.1\)
\(n_{Fe\left(pư\right)}=0.2\left(mol\right)< 0.3\Rightarrow Fedư\)
\(V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{Fe\left(dư\right)}=\left(0.3-0.2\right)\cdot56=5.6\left(g\right)\)
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)
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)
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 \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)
\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\\ n_{O_2}=\dfrac{7,437}{24,79}=0,3mol\\3 Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ \Rightarrow\dfrac{0,3}{3}< \dfrac{0,3}{2}\Rightarrow O_2.dư\\ n_{Fe_3O_4}=0,3:3=0,1mol\\ m_{Fe_3O_4}=232.0,1=23,2g\)