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PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a, \(n_{H_2}=\dfrac{0,7437}{24,79}=0,03\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,06}{0,1}=0,6\left(l\right)\)
b, \(H=\dfrac{669,33}{743,7}.100\%=90\%\)
\(n_{CH_4}=\dfrac{m_{CH_4}}{M_{CH_4}}=\dfrac{1,6}{16}=0,1mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,1 0,2 0,1 ( mol )
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
0,1 0,1 ( mol )
\(m_{CaCO_3}=n_{CaCO_3}.M_{CaCO_3}=0,1.100=10g\)
\(V_{O_2}=n_{O_2}.22,4=0,2.22,4=4,48l\)
\(V_{kk}=V_{O_2}.5=4,48.5=22,4l\)
nCH4 = 1,6/16 = 0,1 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,1 ---> 0,2 ---> 0,1
CO2 + Ca(OH)2 -> CaCO3 + H2O
Mol: 0,1 ---> 0,1 ---> 0,1
mCaCO3 = 0,1 . 100 = 10 (g)
Vkk = 0,2 . 5 . 22,4 = 22,4 (l)
2KMnO4 --to--> MnO2 + O2 + K2MnO4
0,6 <------------------------- 0,3 (mol)
a) nO2 = V/22,4 = 6,12/22,4 ≃ 0,3 (mol)
=> mKMnO4 = n . M = 0,6 . 158 = 94,8 ( g)
b) *PT (a) thu được khí O2
3O2 + 4Al --to--> 2Al2O3
0,3 -> 0,4 (mol)
mO2 = 0,3 . 32 = 9,6 (g)
mAl = 0,4 . 27 = 10,8 (g)
Khối lượng chất rắn cần tìm:
mAl2O3 = mO2 + mAl = 9,6 + 10,8 = 20,4 (g)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
0,2-->0,15
=> V = 0,15.22,4 = 3,36 (l)
b)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,3<----------------------------0,15
=> mKMnO4(lý thuyết) = 0,3.158 = 47,4 (g)
=> \(m_{KMnO_4\left(tt\right)}=\dfrac{47,4.110}{100}=52,14\left(g\right)\)
a)\(n_{Al}=\dfrac{21,6}{27}=0,8\left(m\right)\)
\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
tỉ lệ :4 3 2
số mol :0,8 0,6 0,4
\(V_{O_2}=0,6.22,4=13,44\left(g\right)\)
b)\(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)
c)\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
tỉ lệ : 2 2 3
số mol :0,4 0,4 0,6
\(m_{KClO_3}=0,4.122,5=49\left(g\right)\)
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,4}{2}\) \(\Rightarrow\) Oxi còn dư, Fe p/ứ hết
\(\Rightarrow n_{O_2\left(dư\right)}=0,4-\dfrac{2}{15}=\dfrac{4}{15}\left(mol\right)\)
+) Theo PTHH: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{2}{15}\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{15}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{kk}=\dfrac{2}{15}\cdot22,4\cdot5\approx14,93\left(l\right)\\m_{Fe_3O_4}=\dfrac{1}{15}\cdot232\approx15,47\left(g\right)\end{matrix}\right.\)
a. \(n_{KMnO_4}=\dfrac{47.4}{158}=0,3\left(mol\right)\)
PTHH : 2KMnO4 ---to----> K2MnO4 + MnO2 + O2
0,3 0,15
\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)
b. PTHH : 4Al + 3O2 -> 2Al2O3
0,2 0,15
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)
Pt : \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)
0,02-->0,015-->0,01
a) \(m_{Al2O3}=0,01.102=1,02\left(g\right)\)
b) \(V_{O2\left(dktc\right)}=0,015.24,79=0,37185\left(l\right)\)
sửa lại \(V_{\left(dktc\right)}-->V_{\left(dkc\right)}\)
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, \(H=\dfrac{18,36}{20,4}.100\%=90\%\)
\(\veebar\)