1/
\(n_{O_2}=\dfrac{5}{22,4}=\dfrac{25}{112}\left(mol\right)\)
\(\rightarrow m_{O_2}=\dfrac{25}{112}.32\approx7,14\left(g\right)\)
2/
\(n_{Al}=\dfrac{54}{27}=2\left(mol\right)\)
a/
\(4Al+3O_2\rightarrow2Al_2O_3\)
2 1,5 1 (mol)
b/
\(m_{Al_2O_3}=1.102=102\left(g\right)\)
c/
\(V_{O_2}=22,4.1,5=33,6\left(l\right)\)
Bài 1:
\(n_{o_2}=\dfrac{V}{22,4}=\dfrac{5}{22,4}=\dfrac{25}{112}\left(mol\right)\)
\(\Rightarrow m_{O_2}=n.M=\dfrac{25}{112}.32\approx7,14\left(g\right)\)
Bài 2:
a, PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)
b, \(m_{Al_2O_3}=n.M=1.102=102\left(g\right)\)
c, \(V=22,4.n=22,4.1,5=33,6\left(l\right)\)
Bài 1:
\(n_{O_2}=\dfrac{5}{22,4}=\dfrac{25}{112}\left(mol\right)\)
\(\Rightarrow m_{O_2}=\dfrac{25}{112}\times32=7,14\left(g\right)\)
Bài 2:
a) 4Al + 3O2 \(\underrightarrow{to}\) 2Al2O3
b) \(n_{Al}=\dfrac{54}{27}=2\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}\times2=1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=1\times102=102\left(g\right)\)
c) Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}\times2=1,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,5\times22,4=33,6\left(l\right)\)
1.nO2=\(\dfrac{5}{22.4}\)=0.223mol
mO2=n*M=0.223*32=7.14g
2.a)4Al+3O2➞2Al2O3
2.b)nAl=\(\dfrac{m}{M}\)=\(\dfrac{56}{27}\)=2.074mol
Theo PT: nAl2O3=\(\dfrac{1}{2}\)nAl=\(\dfrac{1}{2}\)*2.074=1.037mol
mAl2O3=n*M=1.037*102=105.774g
2.c) Theo PT: nO2=\(\dfrac{3}{4}\)nAl=\(\dfrac{3}{4}\)*2.074=1.555mol
VO2=n*22.4=1.555*22.4=34.832l (đktc)
BT1: nO2 = \(\dfrac{V}{22,4}\) = \(\dfrac{5}{22,4}\) = \(\dfrac{25}{112}\) (mol)
mO2 = n . M = \(\dfrac{25}{112}\) . 32 \(\approx\) 7,14 (g)
BT2:
A. PTHH: 4Al + 3O2 → 2Al2O3
.....................4........3.............2....... (mol)
.....................2.......1,5............1....... (mol)
B. nAl = \(\dfrac{m}{M}\) = \(\dfrac{54}{27}\) = 2 (mol)
mAl2O3 = n . M = 1 . 102 = 102 (g)
C. VO2 = n . 22,4 = 1,5 . 22,4 = 33,6 (l)
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