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3Fe+2O2-to>Fe3O4
0,225--0,15
n Fe=\(\dfrac{12,6}{56}\)=0,225 mol
VO2=0,15.22,4=3,36l
2KClO3-to>2KCl+3O2
0,1---------------------0,15
=>m KClO3=0,1.122,5=12,25g
\(a,3Fe+2O_2\rightarrow Fe_3O_4\)
\(b,\)
Ta có : \(n_{Fe}=\dfrac{m}{M}=\dfrac{126}{56}=2,25\left(mol\right)\)
\(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.2,25=1,5\left(mol\right)\)
\(\Rightarrow VO_2=33,6\left(l\right)\)
\(c,\)
\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)
Theo \(PTHH:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.1,5=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=n.M=1,122,5=122,5\left(g\right)\)
a)\(n_{Fe}=\dfrac{22,4}{56}=0,4mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,4 \(\dfrac{4}{15}\) \(\dfrac{2}{15}\)
\(V_{O_2}=\dfrac{4}{15}\cdot22,4=5,973l\)
b)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{8}{45}\) \(\dfrac{4}{15}\)
\(m_{KClO_3}=\dfrac{8}{45}\cdot122,5=21,78g\)
\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\
pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
2,25 1,5
=> \(V_{O_2}=1,5.22,4=33,6\left(L\right)\)
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
1 1,5
=> \(m_{KClO3}=122,5\left(g\right)\)
a. \(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
PTHH : 3Fe + 2O2 ---to---> Fe3O4
0,2 \(\dfrac{0.4}{3}\)
b. \(V_{O_2}=\dfrac{0.4}{3}.22,4=\dfrac{8.96}{3}\left(l\right)\)
c. PTHH : 2KClO3 -> 2KCl + 3O2
\(\dfrac{0.8}{3}\) \(\dfrac{0.4}{3}\)
\(m_{KClO_3}=\dfrac{0.8}{3}.122,5=\dfrac{98}{3}\left(g\right)\)
\(n_P=\dfrac{7,44}{31}=0,24mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,24 0,3 0,12
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,2 0,3
\(m_{KClO_3}=0,2\cdot122,5=24,5g\)
a) PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)
b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\) \(\Rightarrow n_{O_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
c) PTHH: \(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,1\left(mol\right)\) \(\Rightarrow m_{KClO_3}=0,1\cdot122,5=12,25\left(g\right)\)
a) PTHH: 4Al + 3O2 =(nhiệt)=> 2Al2O3
nAl = \(\frac{5,4}{27}=0,2\left(mol\right)\)
b) nO2 = \(\frac{0,2\times3}{4}=0,15\left(mol\right)\)
=> VO2(đktc) = 0,15 x 22,4 = 3,36 lít
c) nAl2O3 = \(\frac{0,2\times2}{4}=0,1\left(mol\right)\)
=> mAl2O3 = 0,1 x 102 = 10,2 gam
Câu 3.
a)\(n_{Fe}=\dfrac{12,6}{56}=0,225mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,225 0,15 0,075
\(V_{O_2}=0,15\cdot22,4=3,36l\)
b)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,1 0,15
\(m_{KClO_3}=0,1\cdot122,5=12,25g\)
c)\(n_S=\dfrac{12,8}{32}=0,4mol\)
\(S+O_2\underrightarrow{t^o}SO_2\)
0,4 0,15 0
0,15 0,15 0,15
0,25 0 0,15
\(m_{Sdư}=0,25\cdot32=8g\)
0,4 0,15 0
0,15 0,15 0,15
0,25 0 0,15
là sao?