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PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a, Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
b, Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,075\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,075.232=17,4\left(g\right)\)
c, PT: \(2KMnO_4\underrightarrow{t^o}KMnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
Bạn tham khảo nhé!
4P+5O2-to>2P2O5
0,04----0,05----0,02
n P=0,04 mol
=>m P2O5=0,02.142=2,84g
=>VO2=0,05.22,4=1,12l
c)
2KMnO4-to>K2MnO4+MnO2+O2
0,1----------------------------------------0,05
H=10%
m KMnO4=0,1.158.110%=17,28g
\(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{T^O}2P_2O_5\)
0,04 0,05 0,02
=> \(\left\{{}\begin{matrix}m_{P_2O_5}=0,02.142=2,84\left(g\right)\\V_{O_2}=0,05.22,4=1,12\left(l\right)\end{matrix}\right.\)
\(pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
=> \(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(m_{KMnO_4\left(d\text{ùng}\right)}=15,8.110\%=17,38\left(g\right)\)
a) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
0,3--->0,2----->0,1
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)
c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)
\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
0,44<------------------------------------0,22
\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)
nFe = 5,04 / 56 = 0,09 ( mol)
3Fe + 2O2 --(t^o)-- > Fe3O4
0,09 0,06 0,03 (mol)
=> mFe3O4 = 0,03 . 232 = 6,9(g)
=> VO2 = 0,06 . 22,4 = 1,344 (l)
=> Vkk = 1,344 . 5 = 6,72(l)
a)\(n_{Al}=\dfrac{5,4}{27}=0,2\left(m\right)\)
\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
tỉ lệ :4 3 2
số mol :0,2 0,15 0,1
\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)
b)\(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
c)\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
tỉ lệ :2 1 1 1
số mol :0,3 0,15 0,15 0,15
\(m_{KMnO_4}=0,3.126=37,8\left(g\right)\)
a.b.\(n_{Mg}=\dfrac{m}{M}=\dfrac{6,4}{24}=\dfrac{4}{15}mol\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
4/15 2/15 ( mol )
\(V_{O_2}=n.22,4=\dfrac{2}{15}.22,4=\dfrac{224}{75}l\)
c.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
4/15 2/15 ( mol )
\(m_{KMnO_4}=n.M=\dfrac{4}{15}.158=\dfrac{632}{15}g\)
nMg = 6,4 : 24= 0,26(mol)
pthh : 2Mg+O2 -t--> 2MgO
0,26 --> 0,13 (mol )
=> VO2(đktc) = 0,13.22,4=2,912(l)
pthh : 2KMnO4-t--> K2MnO4 + MnO2+ O2
0,26<------------------------------0,13(mol)
=> mKMnO4 = 0,26.158= 41,08(g)
PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,5\left(mol\right)\) \(\Rightarrow V_{O_2}0,5\cdot22,4=11,2\left(l\right)\)
a) \(4P+5O_2\underrightarrow{t\text{°}}P_2O_5\)
b)\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
Từ PTHH: \(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\)
\(\Rightarrow\)\(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\)
0,15 0,1 0,05
\(m_{Fe_2O_4}=0,05.232=11,6\left(g\right)\\
V_{O_2}=0,1.11,4=2,24\left(l\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,2 0,1
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe_3O_{\text{ 4}}}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.2.......\dfrac{2}{15}\)
\(V_{O_2}=\dfrac{2}{15}\cdot22.4=2.987\left(l\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(\dfrac{4}{15}..............................\dfrac{2}{15}\)
\(m_{KMnO_4}=\dfrac{4}{15}\cdot158=42.13\left(g\right)\)
a) PTHH: 3 Fe + 2 O2 -to-> Fe3O4
b) nFe=0,2(mol) -> nO2= 2/3. 0,2= 2/15 (mol)
=> V(O2,đktc)=22,4. 2/15 \(\approx\) 2,987(l)
c) 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2/15. 2= 4/15(mol)
=>mKMnO4=4/15 x 158 \(\approx\) 42,133(g)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)
c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4\left(LT\right)}=2n_{O_2}=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow m_{KMnO_4\left(LT\right)}=\dfrac{2}{15}.158=\dfrac{316}{15}\left(g\right)\)
Mà: H% = 85%
\(\Rightarrow m_{KMnO_4\left(TT\right)}=\dfrac{\dfrac{316}{15}}{85\%}\approx24,78\left(g\right)\)