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nFe = 16.8/56 = 0.3 (mol)
3Fe + 2O2 -to-> Fe3O4
0.3......0.2...........0.1
VO2 = 0.2*22.4 = 4.48 (l)
mFe3O4 = 0.1*232 = 23.2 (g)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
\(n_{Fe}=\dfrac{6,8}{56}=0,12mol\)
3Fe + 2O2 \(\underrightarrow{t^o}\) Fe3O4
0,12 0,08 0,04 ( mol )
a, \(V_{O_2}=0,08.22,4=1,792l\)
b, mFe3O4 = 0,04.232 = 9,28g
\(n_{Fe}=\dfrac{6,8}{56}=\dfrac{17}{140}(mol)\\ PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ a,n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{17}{210}(mol)\\ \Rightarrow V_{O_2}=\dfrac{17}{210}.22,4=1,81(g)\\ b,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{17}{420}(mol)\\ \Rightarrow m_{Fe_3O_4}=\dfrac{17}{420}.232=9,39(g)\)
gọi số mol Fe là x
pthh : 3Fe + 2O2 --t---> Fe3O4
x---->\(\dfrac{2}{3}\)x----------> \(\dfrac{x}{3}\)
=> mFe3O4= \(\dfrac{x}{3}\) . 232 = \(\dfrac{232x}{3}\) (G)
=> VO2 = \(\dfrac{2x}{3}\) . 22,4 = \(\dfrac{224x}{15}\) (L)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,3\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KClO_3}=0,4.122,5=49\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
\(a.PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 2 1
0,9 0,6 0,3
\(b.V_{O_2}=n.24,79=0,6.24,79=14,874\left(l\right)\)
\(c.m_{Fe_3O_4}=n.M=0,3.\left(56.3+16.4\right)=69,6\left(g\right)\)
\(d.V_{O_2}=14,874\left(l\right)\\ \Rightarrow n_{O_2}=\dfrac{V}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2 2 3
0,6 0,6 0,9
\(m_{KClO_3}=n.M=0,6.\left(39+35,5+16.3\right)=55,5\left(g\right).\)
a.\(n_{CH_4}=\dfrac{V_{CH_4}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,3 0,6 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,6.22,4=13,44l\)
b.
\(n_P=\dfrac{m_P}{M_P}=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
0,1 0,05 ( mol )
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,05.142=7,1g\)
nFe = 33,6 : 56 = 0,6 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,6--> 0,4------->0,2 (mol)
=> vO2 = 0,4.22,4 = 8,96 (mol)
=> mFe3O4 = 0,2.232 = 46,4 (g)
pthh : 2KClO3 -t--> 2KClO3 + 3O2
0,267<-----------------------0,4(mol)
mKClO3= 0,267 .122,5 = 32,67 (g)
Bài 1:
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{3}< \dfrac{0,1}{2}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,1-\dfrac{1}{15}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\dfrac{1}{30}.32\approx1,067\left(g\right)\\V_{O_2\left(dư\right)}=\dfrac{1}{30}.2,24\approx0,746\left(l\right)\end{matrix}\right.\)
b, Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)
Bài 2:
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232\approx15,467\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{2}{15}.22,4\approx2,9867\left(l\right)\)
c, PT: \(2N_2+5O_2\underrightarrow{t^o}2N_2O_5\)
Ta có: \(n_{N_2}=\dfrac{2,8}{28}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{\dfrac{2}{15}}{5}\), ta được N2 dư.
Theo PT: \(n_{N_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{4}{75}\left(mol\right)\)
\(\Rightarrow m_{N_2O_5}=\dfrac{4}{75}.108=5,76\left(g\right)\)
Bạn tham khảo nhé!
Bài 1 :
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{2.24}{224}=0.1\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(Bđ:0.1......0.1\)
\(Pư:0.1.......\dfrac{1}{15}...\dfrac{1}{30}\)
\(Kt:0........\dfrac{1}{30}....\dfrac{1}{30}\)
\(V_{O_2\left(dư\right)}=\dfrac{1}{30}\cdot22.4=0.747\left(l\right)\)
\(m_{Fe_3O_4}=\dfrac{1}{30}\cdot232=7.73\left(g\right)\)
Bài 2 :
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.2.......0.3.......\dfrac{1}{15}\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.47\left(g\right)\)
\(n_{N_2}=\dfrac{2.8}{28}=0.1\left(mol\right)\)
\(2N_2+5O_2\underrightarrow{t^0}2N_2O_5\)
\(0.12......0.3........0.12\)
\(m_{N_2O_5}=0.12\cdot108=12.96\left(g\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\)
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
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(PTHH:3Fe+2O_2-^{t^o}>Fe_3O_4\)
0,6<--------0,4------>0,2
\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(m_{O_2}=n\cdot M=0,4\cdot32=12,8\left(g\right)\)
\(m_{Fe}=n\cdot M=0,6\cdot56=33,6\left(g\right)\)
Câu nào a câu nào b vậy ạ