Bài 8: Tính khối lượng KClO3 cần thiết để điều chế một lượng oxi đủ để đốt cháy hết:
a. Hỗn hợp 0,5 mol CH4 và 0,25mol H2
b. Hỗn hợp 6,75 gam bột nhôm và9,75 gam bột kẽm
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a) PTHH :
2KClO3 \(\rightarrow\)2KCl + 3O2 (1)
CH4 + 2O2 \(\rightarrow\)CO2 + 2H2O (2)
2H2 + O2 \(\rightarrow\) 2H2O (3)
Theo PT(2) => nO2 = 2.nCH4 = 2 x 0.5 =1(mol)
Theo PT(3) => nO2 = 1/2 x nH2 = 1/2 x 0.25 =0.125(mol)
=> tổng nO2 = 1+ 0.125 =1.125(mol)
Theo PT(1) => nKClO3 = 2/3 . nO2 = 2/3 x 1.125 = 0.75(mol)
=> mKClO3 = n .M = 0.75 x 122.5 =91.875(g)
b) 4Al + 3O2 \(\rightarrow\) 2Al2O3 (4)
2Zn + O2 \(\rightarrow\) 2ZnO (5)
nAl = m : M = 6.75/27=0.25(mol)
nZn = m/M = 9.75/65 =0.15(mol)
Theo PT(4) => nO2 = 3/4 . nAl = 3/4 x 0.25 =0.1875(mol)
Theo PT(5) => nO2 = 1/2 x nZn = 1/2 x 0.15 =0.075(mol)
tổng nO2 = 0.1875 + 0.075 =0.2625(mol)
theo PT(1) => nKClO3 = 2/3 x nO2 = 2/3 x 0.2625 =0.175(mol)
=> mKClO3 = n .M = 0.175 x 122.5 =21.4375(g)
\(n_C=\dfrac{1.2}{12}=0.1\left(mol\right)\\ n_S=\dfrac{4}{32}=0.125\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(C+O_2\underrightarrow{t^0}CO_2\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(\sum n_{O_2}=n_C+n_S=0.1+0.125=0.225\left(mol\right)\)
\(\Rightarrow n_{KMnO_4}=2n_{O_2}=0.225\cdot2=0.45\left(mol\right)\)
\(m_{KMnO_4}=0.45\cdot158=71.1\left(g\right)\)
b, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(n_P=\dfrac{6,2}{31}=0,2mol\\n_{O_2}=\dfrac{0,2.5}{4}=0,25mol \)
\(S+O_2\underrightarrow{t^o}SO_2\)
\(n_S=\dfrac{3,2}{32}=0,1mol\\ n_{O_2}=0,1mol\)
\(C+O_2\underrightarrow{t^o}CO_2\)
\(n_C=\dfrac{2,4}{12}=0,2mol\\ n_{O_2}=0,2mol\\ n_{O_2}\left(tổng\right)=\)
\(0,25+0,1+0,2=0,55mol\\ m_{O_2}\left(trong.hh.B\right)=0,55.32=17,6g\)
a, \(m_{Fe}=0,25.56=14g\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(\Rightarrow n_{O_2}=\dfrac{0,25.2}{3}=0,16mol\\ m_{O_2}=0,16.32=5,12g\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(n_{O_2}=\dfrac{0,25.3}{4}=0,1875mol\\ m_{O_2}=0,1875.32=6g\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
\(n_{O_2}=\dfrac{0,5.1}{2}=0,25mol\\ m_{O_2}=0,25.32=8g\)
\(\Rightarrow m_{O_2}\left(trong.hỗn.hợp.A\right)=\) \(5,12+6+8=19,12g\)
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 ...
nAl= 6,75/27=0,25(mol)
nZn= 9,75/65= 0,15(mol)
4 Al + 3 O2 -to-> 2 Al2O3
0,25____0,1875___0,125(mol)
Zn + 1/2 O2 -to->ZnO
0,15___0,075___0,15(mol)
=> n(O2, tổng)= 0,1875+ 0,075= 0,2625(mol)
PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
0,525<----------------------------------------------0,2625(mol)
=> mKMnO4= 0,525.158= 82,95(g)
=> m=82,95(g)
Bài 1
2KClO3--->2KCl+3O2
Theo pthh
n O2=3/2n KClO3=0,75(mol)
m O2=0,75.32=24(g)
Bài 2
a)2KNO3-->2KNO2+O2
b)n O2=1,68/22,4=0,075(mol)
Theo pthh
n KNO3=2n O2=0,15(mol)
m KNO3=0,15.101=15,15(g)
H=85%
-->m KNO3=12,8775(g)
Bài 3
2KClO3--->2KCl+3O2
2O2+CH4--->CO2+2H2O
O2+2H2--->2H2O
Theo pthh2
n O2=2n CH4=1(mol)
Theo pthh3
n O2=1/2n H2=0,125(mol)
\(\sum n_{O2}\) cần =0,125+1=1,125(mol)
Theo pthh1
n KClO3=2/3 n O2=0,75(mol)
m KClO3=0,75.122,5=91,875(g)
Bài 3
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 :
\(\%m_{Mg} = \dfrac{2,85.24}{2,85.24 + 2,7}.100\% = 96,2\%\\ \%m_{Al} = 100\% - 96,2\% = 3,8\%\)
2Zn+O2-to>2ZnO
0,05--0,025---0,05
n Zn=0,05 mol
=>=>VO2=0,025.22,4=0,56l
2KClO3-to->2KCl+3O2
1\60------------------------0,025
=>m KClO3=\(\dfrac{1}{60}\).122,5=2,041g
\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
0,05 0,025
\(V_{O_2}=0,025\cdot22,4=0,56l\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{1}{60}\) 0,025
\(m_{KClO_3}=\dfrac{1}{60}\cdot122,5=2,042g\)
a) \(CH_4+2O_2\underrightarrow{t\text{°}}CO_2+2H_2O\)(1)
___0,5------>1___________________(mol)
\(2H_2+O_2\underrightarrow{t\text{°}}2H_2O\)(2)
0,25->0,125__0,25____(mol)
\(2KClO_3\underrightarrow{t\text{°}}2KCl+3O_2\)(3)
0,75<---------------0,125+1=1,125(mol)
=> mKClO3= 0,75*122,5=91,875(g)
b) n Al= 6,75/27=0,25(mol)
n Zn= 9,75/65= 0,15 (mol)
\(4Al+2O_2\underrightarrow{t\text{°}}2Al_2O_3\) (1')
0.25-->0.125_________(mol)
\(2Zn+O_2\underrightarrow{t\text{}\text{°}}2ZnO\) (2')
0.15->0.075_________(mol)
\(2KClO_3\underrightarrow{t\text{°}}2KCl+3O_2\)(3')
\(\frac{2}{15}\)<-------------------0.075+0.125=0.2(mol)
m KClO3=\(\frac{2}{15}\)*122,5=\(\frac{49}{3}\) (g)
tại sao câu b chỗ PT Al í lại là 2O2 phải là 3O2 chứ