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Fe3O4
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
c, \(H=\dfrac{18,36}{20,4}.100\%=90\%\)
Bài 1:
a) 2CuFeS2 + \(\dfrac{13}{2}\)O2 --to--> 2CuO + Fe2O3 + 4SO2
b) \(n_{CuFeS_2}=\dfrac{3,68}{184}=0,02\left(mol\right)\)
\(n_{O_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,02}{2}< \dfrac{0,075}{\dfrac{13}{2}}\) => CuFeS2 hết, O2 dư
PTHH: 2CuFeS2 + \(\dfrac{13}{2}\)O2 --to--> 2CuO + Fe2O3 + 4SO2
0,02----->0,065------->0,02---->0,01---->0,04
=> \(\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=0,075-0,065=0,01\left(mol\right)\\n_{SO_2}=0,04\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{0,01}{0,01+0,04}.100\%=20\%\\\%V_{SO_2}=\dfrac{0,04}{0,01+0,04}.100\%=80\%\end{matrix}\right.\)
- \(\left\{{}\begin{matrix}m_{CuO}=0,02.80=1,6\left(g\right)\\m_{Fe_2O_3}=0,01.160=1,6\left(g\right)\end{matrix}\right.\)
=> mrắn = 1,6 + 1,6 = 3,2 (g)
Bài 2:
a)
2CuS + 3O2 --to--> 2CuO + 2SO2
4FeS + 7O2 --to--> 2Fe2O3 + 4SO2
b) Gọi số mol CuS, FeS là a, b (mol)
=> 96a + 88b = 22,8 (1)
\(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
=> a + b = 0,25 (2)
(1)(2) => a = 0,1; b = 0,15
=> \(\left\{{}\begin{matrix}n_{CuO}=0,1\left(mol\right)\\n_{Fe_2O_3}=0,075\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1.80}{0,1.80+0,075.160}.100\%=40\%\\\%m_{Fe_2O_3}=\dfrac{0,075.160}{0,1.80+0,075.160}.100\%=60\%\end{matrix}\right.\)
a) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,3------------------>0,15----->0,45
=> \(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)
b)
PTHH: 2H2 + O2 --to-->2H2O
0,45->0,225
=> \(V_{O_2}=0,225.22,4=5,04\left(l\right)\)
=> Vkk = 5,04 : 20% = 25,2 (l)
Đổi 2,5kg = 2500g
mC = 2500 . (100% - 16%) = 2100 (g)
nC = 2100/12 = 175 (mol)
PTHH: C + O2 -> (t°) CO2
Mol: 175 ---> 175 ---> 175
VO2 = 175 . 22,4 = 3920 (l)
mCO2 = 44 . 175 = 7700 (g)
\(m_C=\dfrac{2,5.\left(100-16\right)}{100}=2,1kg\)
\(m_C=2,1kg=2100g\)
\(n_C=\dfrac{m_C}{M_C}=\dfrac{2100}{12}=175mol\)
\(C+O_2\rightarrow\left(t^o\right)CO_2\)
175 175 175 ( mol )
\(V_{O_2}=n_{O_2}.22,4=175.22,4=3920l\)
\(m_{CO_2}=n_{CO_2}.M_{CO_2}=175.44=7700g\)
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
\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)
a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,15<---0,3<----0,15
b) `m_{O_2} = 0,3.32 = 9,6 (g)`
c) `V_{CH_4} = 0,15.22,4 = 3,36 (l)`
a, \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=11,2\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,225\left(mol\right)\Rightarrow V_{O_2}=0,225.22,4=5,04\left(l\right)\)
c, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)
I. a, 3Fe + 2O2 -> (t°) Fe3O4
b, 2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O
c, 4Fe(NO3)3 -> (t°) 2Fe2O3 + 12NO2 + 3O2
d, 2SO2 + O2 -> (t°, V2O5) 2SO3
II. a, nAl = 6,48/27 = 0,24 (mol)
PTHH: 4Al + 3O2 -> (t°) 2Al2O3
Mol: 0,24 ---> 0,18 ---> 0,12
Vkk = 0,18 . 5 . 22,4 = 20,16 (l)
b, Chất sản phẩm: Al2O3: nhôm oxit
mAl2O3 = 0,12 . 102 = 12,24 (g)
I. a, 3Fe + 2O2 -> (t°) Fe3O4
b, 2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O
c, 4Fe(NO3)3 -> (t°) 2Fe2O3 + 12NO2 + 3O2
d, 2SO2 + O2 -> (t°, V2O5) 2SO3
II. a, nAl = 6,48/27 = 0,24 (mol)
PTHH: 4Al + 3O2 -> (t°) 2Al2O3
Mol: 0,24 ---> 0,18 ---> 0,12
Vkk = 0,18 . 5 . 22,4 = 20,16 (l)
b, Chất sản phẩm: Al2O3: nhôm oxit
mAl2O3 = 0,12 . 102 = 12,24 (g)