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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
4Al + 3O2 --to--> 2Al2O3
2Mg + O2 --to--> 2MgO
b) Gọi số mol Al, Mg là a, b (mol)
=> 27a + 24b = 7,8 (1)
PTHH: 4Al + 3O2 --to--> 2Al2O3
a--->0,75a----->0,5a
2Mg + O2 --to--> 2MgO
b--->0,5b------->b
=> 102.0,5a + 40b = 14,2
=> 51a + 40b = 14,2 (2)
(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
nO2 = 0,75a + 0,5b = 0,2 (mol)
=> VO2 = 0,2.22,4 = 4,48 (l)
=> Vkk = 4,48 : 20% = 22,4 (l)
c)
mAl = 0,2.27 = 5,4 (g)
mMg = 0,1.24 = 2,4 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
4Al + 3O2 --to--> 2Al2O3
2Mg + O2 --to--> 2MgO
b) Gọi số mol Al, Mg là a, b (mol)
=> 27a + 24b = 7,8 (1)
PTHH: 4Al + 3O2 --to--> 2Al2O3
a--->0,75a----->0,5a
2Mg + O2 --to--> 2MgO
b--->0,5b------->b
=> 102.0,5a + 40b = 14,2
=> 51a + 40b = 14,2 (2)
(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
nO2 = 0,75a + 0,5b = 0,2 (mol)
=> VO2 = 0,2.22,4 = 4,48 (l)
=> Vkk = 4,48 : 20% = 22,4 (l)
c)
mAl = 0,2.27 = 5,4 (g)
mMg = 0,1.24 = 2,4 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Bảo toàn khối lượng: \(m_{O_2}=m_{oxit\:}-m_{hh}=6,6\left(g\right)\)
\(\Rightarrow n_{O_2}=\dfrac{6,6}{32}=\dfrac{33}{160}\left(mol\right)\) \(\Rightarrow V_{O_2}=\dfrac{33}{160}\cdot22,4=4,62\left(l\right)\)
Bài 1 :
\(BTKL:\)
\(m_{hh}+m_{O_2}=m_{oxit}\)
\(\Rightarrow10.5+m_{O_2}=17.1\)
\(\Rightarrow m_{O_2}=17.1-10.5=6.6\left(g\right)\)
\(V_{O_2}=\dfrac{6.6}{32}\cdot22.4=4.62\left(l\right)\)
Bài 2 :
\(A:XO_n\)
\(B:H_mY\)
\(\%O=\dfrac{16n}{X+16n}\cdot100\%=50\%\)
\(\Rightarrow X+16n=32n\)
\(\Rightarrow X=16n\)
\(n=2\Rightarrow X=32\)
\(CT:SO_2\)
\(\%H=\dfrac{m}{m+Y}\cdot100\%=25\%\)
\(\Rightarrow Y=3m\left(1\right)\)
\(Mà:\) \(\dfrac{M_{SO_2}}{M_B}=4\)
\(\Leftrightarrow M_B=\dfrac{64}{4}=16\)
\(\Leftrightarrow Y+m=16\left(2\right)\)
\(\left(1\right),\left(2\right):\)
\(\left\{{}\begin{matrix}Y=12\\m=4\end{matrix}\right.\)
\(CT:CH_4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,1<-0,05<-------0,1
2CO + O2 --to--> 2CO2
0,2<--0,1-------->0,2
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)
\(2CO+O_2\rightarrow2CO_2\)
a 0,5a a
\(2H_2+O_2\rightarrow2H_2O\)
0,1 0,05 \(\leftarrow\) 0,1
\(\Sigma n_{O_2}=0,5a+0,05=0,15\)
\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)
\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{CO_2}=0,2\cdot44=8,8g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\) (1)
\(4H_2+O_2\underrightarrow{t^o}2H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)
c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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, Vì đốt cháy X trong khí O2 thu được CO2 và H2O nên X chắc chắn có nguyên tố C và H, có thể có O.
b, Theo ĐLBT KL, có: mX + mO2 = mCO2 + mH2O
⇒ mO2 = mCO2 + mH2O - mX = 47,49 + 21,42 - 15,46 = 53,45 (g)
\(\Rightarrow n_{O_2}=\dfrac{53,45}{32}1,6703125\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,6703125.22,4=37,415\left(l\right)\)
Bạn tham khảo nhé!