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a.\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,6 0,45 0,3 ( mol )
\(m_{Al}=0,6.27=16,2g\)
\(V_{O_2}=0,45.22,4=10,08l\)
\(V_{kk}=10,08.5=50,4l\)
b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,3 0,45 ( mol )
\(m_{KClO_3}=0,3.122,5=36,75g\)
c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,3 0,45 ( mol )
\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)
\(m_{KClO_3}=0,4.122,5=49g\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
nAl=16,2/27= 0,6(mol)
a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3
nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)
=> V(O2,đktc)=0,45 x 22,4=10,08(l)
b) nAl2O3= nAl/2=0,6/2=0,3(mol)
=>mAl2O3=102. 0,3= 30,6(g)
c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2=2. 0,45=0,9(mol)
=>mKMnO4= 158 x 0,9= 142,2(g)
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
\(m_{KClO_3}=0,4.122,5=49\left(g\right)\)
mMg = 3.6/24 = 0.15 (mol)
2Mg + O2 -to-> 2MgO
0.15__0.075____0.15
mMgO= 0.15*40 = 6 (g)
VO2 = 0.075*22.4 = 1.68 (l)
2KClO3 -to-> 2KCl + 3O2
0.05_______________0.075
mKClO3 = 0.05*122.5 = 6.125 (g)
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)
c) PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)
4Al+3O2-to>2Al2O3
0,4----0,3-----0,2
n Al=0,4 mol
=>m Al2O3=0,2.102=20,4g
=>VO2=0,3.22,4=6,72l
2KClO3-to>2KCl+3O2
0,2----------------------0,3
=>m KClO3=0,2.122,5=24,5g
nAl = 10,8 : 27 = 0,4 (mol)
pthh : 4Al + 3O2 -t--> 2Al2O3
0,4-->0,3-------> 0,2 (mol)
mAl2O3 = 0,2 . 102 = 20,4 (g)
VH2 = 0,3 . 22,4 = 6,72 (L)
pthh: 2KClO3 -t--> 2KCl + 3O2
0,2<----------------------0,3 (mol)
=> mKClO3 = 0,2 . 122,5 = 24,5 (g)