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a. \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)
PTHH : 3Fe + 2O2 -to-> Fe3O4
0,09 0,06 0,03
\(m_{Fe}=0,09.56=5,04\left(g\right)\)
\(V_{O_2}=0,06.22,4=1,344\left(l\right)\)
b. PTHH : 2KCl + 3O2 -> 2KClO3
0,06 0,04
\(m_{KClO_3}=0,04.122,5=4,9\left(g\right)\)
nFe3O4 = 17.4/232 = 0.075 (mol)
3Fe + 2O2 -to-> Fe3O4
0.225__0.15_____0.075
mFe = 0.225*56=12.6 (g)
VO2 = 0.15*22.4 = 3.36 (l)
2KClO3 -to-> 2KCl + 3O2
0.1________________0.15
mKClO3 = 0.1*122.5 = 12.25 (g)
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?
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. \(n_{Fe_3O_4}=\dfrac{2.32}{232}=0,01\left(mol\right)\)
\(3Fe+2O_2\rightarrow Fe_3O_4\)
0,03 ..0,02 ...... 0,01 (mol)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,03.56=1,68\left(g\right)\\V_{O_2}=0,02.22,4=0,448\left(l\right)\end{matrix}\right.\)
\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
0,04 ------------------------------------ 0,02 (mol)
\(\rightarrow m_{KMnO_4}=0,04.158=6,32\left(g\right)\)
a)PTHH: 3Fe+2O2__>Fe3O4
SỐ MOL OXIT SẮT TỪ:2.32/232=0.1MOL
THEO PT:
- ĐỂ TẠO RA 1 MOL Fe3O4 CẦN 2 MOL O2 VẬY ĐỂ TẠO RA 0.1 MOL Fe3O4 CẦN 0.2 MOL O2
- ĐỂ TẠO RA 1 MOL Fe3O4 CẦN 3 MOL Fe VẬY ĐỂ TẠO RA 0.1 MOL Fe3O4 CẦN 0.3 MOL Fe
=>SỐ GAM Fe CẦN: 0.3x56=16.8 g
-THỂ TÍCH KHÍ O2 CẦN: 0.2x22.4=4.48 l
b) PTHH: 2KMnO4 __>MnO2+O2+K2MnO4
Theo PT,để tạo 1 mol O2 cần 2 mol KMnO4
⇒để tạo ra 0.2 mol O2(cần dùng) thì cần 0.4 mol KMnO4
⇒ số g KMnO4 cần là:158*0.4=63.2
P/S: đúng thì nhớ tích vào đấy
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 5:
a,\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,03 0,02 0,01
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,03.56=1,68\left(g\right)\\m_{O_2}=0,02.32=0,64\left(g\right)\end{matrix}\right.\)
b, \(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,04 0,02
\(\rightarrow m_{KMnO_4}=0,04.158=6,32\left(g\right)\)
Bài 6:
a, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
LTL: \(\dfrac{0,2}{4}< \dfrac{0,3}{5}\) => O2 dư
\(\rightarrow\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}V_{O_2\left(dư\right)}=\left(0,3-0,25\right).22,4=1,12\left(l\right)\\m_{P_2O_5}=0,1.142=14,2\left(g\right)\end{matrix}\right.\)
Bạn tách ra từng câu nhé!
Bài 3.
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{36}{56}=0,6428mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,6428 ----- 0,4285 ( mol )
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,857 0,4285 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,857.158=135,406g\)
Bài 4.
a.\(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{51}{102}=0,5mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
1 0,75 0,5 ( mol )
\(m_{Al}=n_{Al}.M_{Al}=1.27=27g\)
\(V_{O_2}=n_{O_2}.22,4=0,75.22,4=16,8l\)
b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
1,5 0,75 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=1,5.158=237g\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
0,5 0,75 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,5.122,5=61,25g\)
a, Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
THeo PT: \(n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\Rightarrow V_{O_2}=0,04.22,4=0,896\left(l\right)\)
b, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=0,08\left(mol\right)\Rightarrow m_{KMnO_4}=0,08.158=12,64\left(g\right)\)
a) \(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02\left(mol\right)\).
PTHH : \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Mol : 3 : 2 : 1
Mol 0,04 ← 0,02
\(\Rightarrow V_{O_2}=n_{O_2}.22,4=\left(0,04\right).\left(22,4\right)=0,896\left(l\right)\).
b) Từ phương trình ở câu a \(\Rightarrow n_{O_2}=0,04\left(mol\right)\).
PTHH : \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Mol : 2 : 1 : 1 : 1
Mol : 0,08 ← 0,04
\(\Rightarrow m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\left(0,08\right).158=12,64\left(g\right)\).
nFe3O4 = 4.64/232 = 0.02 (mol)
3Fe + 2O2 -to-> Fe3O4
0.06___0.04_____0.02
mFe = 0.06*56 = 3.36 (g)
mO2 = 0.04*32 = 1.28 (g)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.08_________________________0.04
mKMnO4 = 0.08*158 = 12.64 (g)
Chúc em học tốt !!!