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\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
tỉ lệ 2 : 2 : 3
n(mol) `1/3`<------------`1/3`<-----`0,5`
\(m_{KClO_3}=n\cdot M=\dfrac{1}{3}\cdot\left(39+35,5+16\cdot3\right)\approx40,83\left(g\right)\)
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6mol\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,4 0,6 ( mol )
\(m_{KClO_3}=0,4.122,5=49g\)
a.\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)
0,1 0,05 ( mol )
\(V_{kk}=\left(0,05.22,4\right).5=5,6l\)
b.\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
1/30 0,05 ( mol )
\(m_{KClO_3}=\dfrac{1}{30}.122,5=4,08g\)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_{O_2}=2n_{Fe_3O_4}=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
\(m_{O_2}=0,04.32=1,28\left(g\right)\)
b, Phần này đề bài cho là KMnO4 hay KClO3 vậy bạn?
1) nO2= 48/32=1,5(mol)
PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4=2.1,5=3(mol)
=>mKMnO4=158.3= 474(g)
2) nO2=2,24/22,4=0,1(mol)
nKMnO4=2.0,1=0,2(mol)
-> mKMnO4=158.0,2= 31,6(g)
\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,03 0,02 0,01 ( mol )
\(m_{Fe_3O_4}=0,01.232=2,32\left(g\right)\)
\(V_{kk}=0,02.22,4.5=2,24\left(l\right)\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
\(\dfrac{1}{75}\) 0,02 ( mol )
\(m_{KClO_3}=\dfrac{1}{75}.122,5=1,63\left(g\right)\)
a,nFe=1,68/56=0,03 mol
Ta có PTHH : 3Fe + 2O2 --> Fe3O4 (1) ( ở trên dấu --> có to nha )
Theo PTHH ta có :
nFe3O4=1/3nFe=1/3.0,03=0,01 mol
nO2=2/3nFe=2/3.0,03=0,02 mol
=>mFe3O4= 0,01.232=2,32g
=>Vkk=5.(0,02.22,4)=2,24 l
b, Ta có PTHH: 2KClO3 --> 2KCl + 3O2 (2) ( trên dấu --> vẫn có to )
Gọi x là số mol KClO3 cần dùng ( x > 0 )
Theo PTHH (3) và theo bài ra ta có PTHH sau:
2/3x=0,02
=> x=0,03 mol
=> mKClO3= 0,03.122,5= 3,675g
a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
b) \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)
c) \(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)
d) \(m_{O_2}=\dfrac{4,958.0,99}{0,082.\left(273+25\right)}=0,2\left(mol\right)\)
e) \(m_{CH_4}=\dfrac{12,359.0,99}{0,082\left(273+25\right)}=0,5\left(mol\right)\)
a: \(n=\dfrac{28}{56}=0.5\left(mol\right)\)
b: \(n=\dfrac{13.5}{27}=0.5\left(mol\right)\)
\(a)\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3.2 = 0,6(mol)\\ \Rightarrow m_{KMnO_4} = 0,6.158 = 94,8(gam)\\ b)\ n_{KClO_3} = \dfrac{24,5}{122,5} = 0,2(mol)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,3(mol)\\ \Rightarrow m_{O_2} = 0,3.32 = 9,6(gam)\)
a) nO2 = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Pt: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Theo pt: \(n_{KMnO_4}=2nO_2=0,6\left(mol\right)\Rightarrow m_{KMnO_4}=0,6.158=94,8g\)
b) \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,3\left(mol\right)\)
\(Pt:2KClO_3\rightarrow2KCl+3O_2\)
\(nO_2=\dfrac{3}{2}n_{KClO_3}=0,45\left(mol\right)\Rightarrow mO_2=0,45.32=14,4g\)
1. \(n_{O_2}=\frac{V}{22,4}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
2.
\(n_{CO_2}=\frac{m}{M}=\frac{4,4}{44}=0,1\left(mol\right)\)
\(n_{O_2}=\frac{m}{M}=\frac{3,2}{32}=0,1\left(mol\right)\)
\(V_{HC}=n.22,4=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)
\(B1\\ n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\\ m_{KClO_3}=122,5.\dfrac{1}{3}=\dfrac{245}{6}\left(g\right)\\ B2:n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)=n_{O_2\left(bài1\right)}\\ \Rightarrow n_{KClO_3}=\dfrac{1}{3}\left(mol\right)\\ m_{KClO_3}=\dfrac{245}{6}\left(g\right)\)