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\(a) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{Al} = \dfrac{5,4}{27} = 0,2(mol)\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,1(mol) \Rightarrow m_{Al_2O_3} = 0,1.102 = 10,2(gam)\\ b) n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3(mol) \Rightarrow m_{KMnO_4} = 0,3.158 = 47,4(gam)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,2.158=47,4\left(g\right)\)
Bạn tham khảo nhé!
\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ Mol:2,25\rightarrow1,5\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ Mol:1\leftarrow1\leftarrow1,5\\ m_{KClO_3}=1.122,5=122,5\left(g\right)\)
nFe= 2,8/56=0,05(mol)
PTHH: 3Fe+2O2-> Fe3O4
0,05 0,003 0,0167
a) mFe3O4= 0,0167. 232=3,8744(g)
b) Vo2 = 0,003.22,4=0,0672(l)
Ta có : nFe=2,8/56=0,05 mol
3Fe+2O2--->Fe3O4 (1)
a) theo pt (1) nFe3O4=1/3nFe=1/60 mol
=> mFe3O4=1/60 . MFe3O4 =58/15 gam
b) nO2 =2/3 nFe=0,05 .2/3=1/30 mol
=> VO2=1/30 .22,4=0,7467 lít
c) Vkhông khí=5. VO2=3,73 lít.
PTHH: 3Fe + 2O2 -to-> Fe3O4
Ta có:
\(n_{Fe_3O_4}=\frac{2,32}{232}=0,01\left(mol\right)\)
a) Theo PTHH và đề bài, ta có:
\(n_{Fe}=3.0,01=0,03\left(mol\right)\\ n_{O_2}=2.0,01=0,02\left(mol\right)\)
Khối lượng sắt cần dùng:
\(m_{Fe}=0,03.56=1,68\left(g\right)\)
Thể tích khí O2 tham gia (đktc):
\(V_{O_2\left(đktc\right)}=0,02.22,4=0,448\left(l\right)\)
c) PTHH: 2KClO3 -to-> 2KCl + 3O2
Ta có: \(n_{O_2}=0,02\left(mol\right)\)
=> \(n_{KClO_3}=\frac{3.0,02}{2}=0,03\left(mol\right)\)
Khối lượng KClO3 cần dùng:
\(m_{KClO_3}=0,03.122,5=3,675\left(g\right)\)
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(n_{Fe_3O_4}=\frac{2,32}{232}=0,01mol\)
\(n_{Fe}=3.n_{Fe_3O_4}=3.0,01=0,03mol\)
\(n_{O_2}=2.n_{Fe_3O_4}=2.0,01=0,02mol\)
\(m_{Fe}=0,03.56=1,68g\)
\(V_{O_2}=0,02.22,4=0,448l\)
b) Gợi ý: Viết PTHH
Tính số mol KClO3 dựa theo số mol oxi
Nhân cho khối lượng mol tính khối lượng
a, 3Fe + 2O2 -to-> Fe3O4
b, nFe = m/M = 16,8/56 = 0,3 (mol)
từ pthh ta có: \(n_{O_2}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)
=>\(V_{O_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)
c, C1: từ pthh ta có: \(n_{Fe_3O_4}=\dfrac{0,3.1}{3}=0,1\left(mol\right)\)
=>\(m_{Fe_3O_4}=n.M=0,1.\left(56.3+4.16\right)=0,1.232=23,2\left(g\right)\)
C2: \(m_{O_2}=n.M=0,2.32=6,4\left(g\right)\)
Áp dụng ĐLBTKL ta co:
\(m_{Fe_3O_4}=m_{Fe}+m_{O_2}=16,8+6,4=23,2\left(g\right)\)
Câu 8
a)\(n_{Fe3O4}=\frac{23,2}{232}=0,1\left(mol\right)\)
\(3Fe+2O2-->Fe3O4\)
0,3-----0,2----------------0,1(mol)
\(m_{Fe}=0,3.56=16,8\left(g\right)\)
\(m_{O2}=0,2.32=6,4\left(g\right)\)
b)\(2KMnO4-->K2MnO4+MnO2+O2\)
\(n_{KMnO4}=2n_{O2}=0,4\left(mol\right)\)
\(m_{KMnO4}=0,4.158=63,2\left(g\right)\)
Bài 9
a)\(4Al+3O2-->2Al2O3\)
\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
\(n_{O2}=\frac{3}{2}n_{Al2O3}=0,3\left(mol\right)\)
\(V_{O2}=0,3.22,4=6,72\left(l\right)\)
b)\(2KMnO4-->K2MnO4+MnO2+O2\)
\(n_{KMnO4}=2n_{O2}=0,6\left(mol\right)\)
\(m_{KMnO4}=0,6.158=94,8\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\a, PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ b,n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3.0,2}{4}=0,15\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=2.n_{O_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow m_{KMnO_4}=158.0,3=47,4\left(g\right)\)
a: PTHH: 3Fe + 2O2 --> Fe3O4
nFe3O4 = 3,48/232 = 0,015 mol
Theo PTHH : nFe = 3nFe3O4 = 3.0,15 = 0,045 mol
=>mFe = 0,045.56 = 2,52 g
Theo PTHH nO2 = 2nFe3O4 = 0,015.2 = 0,03 mol
=> VO2 = 0,03.22,4 = 0,672 lít
b: PTHH: 2KClO3 to---> 2KCl + 3O2
Theo PTHH nKClO3 = 2/3nO2 = 2/3.0,03 = 0,02 mol
=> mKClO3 = 0,02.122,5=2,45g
chúc bạn học tốt :))
a ) PTHH : 3Fe + 2O2 \(\rightarrow\)Fe3O4
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{3,48}{232}=0,015\left(mol\right)\)
Theo PTHH , ta có :
* \(n_{Fe}=n_{Fe_3O_4}\cdot3=0,015\cdot3=0,045\left(mol\right)\)
\(\Rightarrow m_{Fe}=n\cdot M=0,045\cdot56=2,52\left(g\right)\)
* \(n_{O_2}=n_{Fe_3O_4}\cdot2=0,015\cdot2=0,03\left(mol\right)\)
\(\Rightarrow V_{O_2}=n\cdot22,4=0,03\cdot22,4=0,672\left(l\right)\)
b ) PTHH : 2KClO3 \(\rightarrow\)2KCl + 3O2
Theo PTHH , ta có :
\(n_{KClO_3}=n_{O_2}\cdot\dfrac{2}{3}=0,03\cdot\dfrac{2}{3}=0,02\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=n\cdot M=0,02\cdot122,5=2,45\left(g\right)\)