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a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
\(m_{O_2}=0,04.32=1,28\left(g\right)\)
b, Phần này đề bài cho là KMnO4 hay KClO3 vậy bạn?
a, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)
b, Ta có: \(n_{Cu}=\dfrac{16,8}{64}=0,2625\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{1}{2}n_{Cu}=0,13125\left(mol\right)\\n_{CuO}=n_{Cu}=0,2625\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{O_2}=0,13125.32=4,2\left(g\right)\)
\(m_{CuO}=0,2625.80=21\left(g\right)\)
c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2625\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,2625.158=41,475\left(g\right)\)
Bạn tham khảo nhé!
a)\(n_{Fe_3O_4}=\dfrac{11,6}{232}=0,05mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe}=0,15\cdot56=8,4g\)
\(m_{O_2}=0,1\cdot32=3,2g\)
b)\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,2 0,1
\(m_{KMnO_4}=0,2\cdot158=31,6g\)
\(a,n_{Fe_3O_4}=\dfrac{11,6}{232}=0,05\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,15<--0,1<----------0,05
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,15.56=8,4\left(g\right)\\m_{O_2}=0,1.32=3,2\left(g\right)\end{matrix}\right.\)
b, PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,2<--------------------------------------0,1
=> mKMnO4 = 0,2.158 = 31,6 (g)
a,\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,03<--0,02<-------0,01
\(\left\{{}\begin{matrix}m_{Fe}=0,03.56=1,68\left(g\right)\\m_{O_2}=0,02.32=0,64\left(g\right)\end{matrix}\right.\)
b, PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,04<-----------------------------------0,02
\(m_{KMnO_4}=\dfrac{0,04.158}{85\%}=7,435\left(g\right)\)
\(n_{Fe_3O_4}=\dfrac{2.32}{232}=0.01\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{t^0}}Fe_3O_4\)
\(0.03......0.02.........0.01\)
\(m_{Fe}=0.03\cdot56=1.68\left(g\right)\)
\(m_{O_2}=0.02\cdot32=0.64\left(g\right)\)
\(2KMnO_4\underrightarrow{^{t^0}}K_2MnO_4+MnO_2+O_2\)
\(0.04............................................0.02\)
\(m_{KMnO_4}=0.04\cdot158=6.32\left(g\right)\)
a)
n Fe3O4 = 2,32/232 = 0,01(mol)
3Fe + 2O2 \(\xrightarrow{t^o}\) Fe3O4
0,03....0,02.......0,01...........(mol)
m Fe = 0,03.56 = 1,68(gam)
m O2 = 0,02.32= 0,64(gam)
c)
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
n KMnO4 = 2n O2 = 0,04(mol)
m KMnO4 = 0,04.158 = 6,32 gam
Sửa đề: 4,46 (g) → 4,64 (g)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo PT: \(n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
\(n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\Rightarrow m_{O_2}=0,04.32=1,28\left(g\right)\)
b, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,08\left(mol\right)\Rightarrow m_{KMnO_4}=0,08.158=12,64\left(g\right)\)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
Theo PT: \(n_{Fe}=3n_{Fe_3O_4}=0,03\left(mol\right)\Rightarrow m_{Fe}=0,03.56=1,68\left(g\right)\)
\(n_{O_2}=2n_{Fe_3O_4}=0,02\left(mol\right)\Rightarrow m_{O_2}=0,02.32=0,64\left(g\right)\)
b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,04\left(mol\right)\Rightarrow m_{KMnO_4}=0,04.158=6,32\left(g\right)\)
Bài 3 :
PTHH : \(6Fe+4O_2\left(t^o\right)->2Fe_3O_4\) (1)
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{2,32}{56.3+16.4}=0,01\left(mol\right)\)
Từ (1) => \(3n_{Fe_3O_4}=n_{Fe}=0,03\left(mol\right)\)
=> \(m_{Fe}=n.M=1,68\left(g\right)\)
Từ (1) => \(2n_{Fe_3O_4}=n_{O_2}=0,02\left(mol\right)\)
=> \(V_{O_2\left(đktc\right)}=n.22,4=0,448\left(l\right)\)
Bài 4 :
PTHH : \(4P+5O_2\left(t^o\right)->2P_2O_5\) (1)
\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{32}=0,21\left(mol\right)\)
Có : \(n_P< n_{O_2}\left(0,2< 0,21\right)\)
-> P hết ; O2 dư
Từ (1) -> \(\dfrac{1}{2}n_P=n_{P_2O_5}=0,1\left(mol\right)\)
=> \(m_{P_2O_5}=n.M=14,2\left(g\right)\)
Bài 3:
\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
PTHH: 3Fe + 2O2 ---to→ Fe3O4
Mol: 0,03 0,02 0,01
\(m_{Fe}=0,03.56=1,68\left(g\right);V_{O_2}=0,02.22,4=0,448\left(l\right)\)
a, \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)
PTHH: 3Fe + 2O2 ----to----> Fe3O4
Mol: 0,09 0,06 0,03
\(m_{Fe}=0,09.56=5,04\left(g\right)\)
\(m_{O_2}=0,06.32=1,92\left(g\right)\)
b,
PTHH: 2KClO3 ----to---> 2KCl + 3O2
Mol: 0,02 0,06
\(m_{KClO_3}=0,02.122,5=2,45\left(g\right)\)