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PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\) (1)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (2)
Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\Rightarrow n_{O_2\left(2\right)}=0,075\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(1\right)}=1,425\left(mol\right)\) \(\Rightarrow n_{Mg}=2,85\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{2,85\cdot24}{2,85\cdot24+2,7}\cdot100\%\approx96,2\%\)
\(\Rightarrow\%m_{Al}=3,8\%\)
\(n_{O_2} =\dfrac{33,6}{22,4} = 1,5(mol)\\ n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{O_2} = \dfrac{1}{2}n_{Mg} + \dfrac{3}{4}n_{Al}\\ \Rightarrow n_{Mg} = 2,85(mol)\)
Vậy :
\(\%m_{Mg} = \dfrac{2,85.24}{2,85.24 + 2,7}.100\% = 96,2\%\\ \%m_{Al} = 100\% - 96,2\% = 3,8\%\)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (1)
\(2Mg+O_2\underrightarrow{t^o}2MgO\) (2)
Ta có: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}\cdot\dfrac{13,5}{27}=0,375\left(mol\right)\\n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(2\right)}=0,375\left(mol\right)\) \(\Rightarrow n_{Mg}=0,75\left(mol\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,75\cdot24}{0,75\cdot24+13,5}\cdot100\%\approx57,14\%\)
Gọi số mol C, S là a, b
=> 12a + 32b = 7,68
PTHH: C + O2 --to--> CO2
_____a--------------->a
S + O2 --to--> SO2
b--------------->b
=> a + b = \(\dfrac{9,856}{22,4}=0,44\)
=> a = 0,32; b = 0,12
=> \(\left\{{}\begin{matrix}\%C=\dfrac{0,32.12}{7,68}.100\%=50\%\\\%S=\dfrac{0,12.32}{7,68}.100\%=50\%\end{matrix}\right.\)
a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)
\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)
\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)
b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)
\(\rightarrow V_{kk}=4,48.5=11,2l\)
c. Có \(n_{Mg}=2n_{O_2}=0,2l\)
\(\rightarrow m_{Mg}=0,2.24=4,8g\)
\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)
\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)
a) Gọi số mol Al, Zn là 2a, a (mol)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2a-->1,5a---------->a
2Zn + O2 --to--> 2ZnO
a---->0,5a------->a
=> \(102a+81a=18,3\)
=> a = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)
b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(2Mg+O_2\xrightarrow{t^o}2MgO\\ 4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ \Rightarrow \begin{cases} 24.n_{Mg}+27.n_{Al}=5,1\\ 0,5.n_{Mg}+0,75.n_{Al}=n_{O_2}=\dfrac{2,8}{22,4}=0,125 \end{cases}\\ \Rightarrow \begin{cases} n_{Mg}=0,1(mol)\\ n_{Al_2O_3}=0,1(mol) \end{cases}\\ \Rightarrow \%m_{Mg}=\dfrac{0,1.24}{5,1}.100\%\approx 47,06\%\)
Gọi nC = a (mol); nS = b (mol)
12a + 32b = 12 (1)
PTHH:
C + O2 -> (t°) CO2
a ---> a ---> a
S + O2 -> (t°) SO2
b ---> b ---> b
44a + 64b = 28 (2)
Từ (1)(2) => a = 0,2 (mol); b = 0,3 (mol)
nO2 = 0,2 + 0,3 = 0,5 (mol)
VO2 = 0,5 . 22,4 = 11,2 (l)
Bài 1 :
\(n_{Na}=\dfrac{m}{M}=0,1\left(mol\right)\)
\(4Na+O_2\rightarrow2Na_2O\)
..0,1....0,025....0,05.......
a, \(V_{O_2}=n.22,4=0,56\left(l\right)\)
b, \(m=m_{Na_2o}=n.M=3,1\left(g\right)\)
Bài 2 :
\(n_{Al}=\dfrac{m}{M}=0,1\left(mol\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\)
..0,1...0,075...
\(\Rightarrow n_{O_2}=0,075\left(mol\right)\)
Mà : \(\Sigma n_{O_2}=\dfrac{V}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow n_{O_2\left(Mg\right)}=0,4-0,075=0,325\left(mol\right)\)
\(2Mg+O_2\rightarrow2MgO\)
.0,65.....0,325........
\(\Rightarrow m_{Mg}=15,6\left(g\right)\)
\(\Rightarrow m_{hh}=2,7+15,6=18,3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Al=~14,75\\\%Mg=~85,25\end{matrix}\right.\) %
Bài 3 :
- Gọi số mol Al và Mg lần lượt là x , y
\(4Al+3O_2\rightarrow2Al_2O_3\)
..x....0,75x
\(2Mg+O_2\rightarrow2MgO\)
..y........0,5y...........
Có : \(n_{O_2}=0,75x+0,5y=\dfrac{V}{22,4}=0,1\left(mol\right)\left(I\right)\)
Lại có : \(m_{hh}=m_{Al}+m_{Mg}=27x+24y=3,9\left(II\right)\)
- Giair ( i ) và ( ii ) ta được : \(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\) ( mol )
\(\Rightarrow\left\{{}\begin{matrix}\%Al=~69,23\\\%Mg=~30,77\end{matrix}\right.\) %
Vậy ...