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a)
nKMnO4 = 47.4/158 = 0.3 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.3_________________________0.15
VO2 = 0.15*22.4 = 3.36 (l)
b)
nKMnO4 = 31.6/158 = 0.2 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.2_________________________0.1
VO2 = 0.1*22.4 = 2.24 (l)
c)
nKMnO4 = 39.5/158 = 0.25 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.25_________________________0.125
VO2 = 0.125*22.4 = 2.8 (l)
2)
a)
nO2 = 3.36/22.4 = 0.15 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.3_________________________0.15
mKMnO4 = 0.3*158 = 47.4(g)
b)
nO2 = 8.96/22.4 = 0.4 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.8_________________________0.4
mKMnO4 = 0.8*158 = 126.4(g)
c)
nO2 = 14.4/32 = 0.45 (mol)
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.9_________________________0.45
mKMnO4 = 0.9*158 = 142.2(g)
\(n_{Mg}=0,1\left(mol\right)\)
\(2Mg+O_2\underrightarrow{^{to}}2MgO\)
\(\Rightarrow n_{O2}=0,05\left(mol\right);n_{MgO}=0,1\left(mol\right)\)
\(\Rightarrow V_{O2}=0,05.22,4=1,12\left(l\right)\)
\(m_{MgO}=0,1.40=4\left(g\right)\)
\(2KClO_3\underrightarrow{^{to}}2KCl+3O_2\)
\(n_{KClO3}=\frac{0,1}{3}\left(mol\right)\)
\(\Rightarrow_{KClO3}=4,08\left(g\right)\)
a, PTHH:
Fe2O3 + 3H2 ---to---> 2Fe + 3H2O (1)
CuO + H2 ---to---> Cu + H2O (2)
b, nFe = \(\dfrac{2,8}{56}=0,05\left(mol\right)\)
nCu = \(\dfrac{6-2,8}{64}=0,05\left(mol\right)\)
Theo pt (1): nH2 (1) = 2nFe = 2 . 0,05 = 0,1 (mol)
Theo pt (2): nH2 (2) = nCu = 0,05 (mol)
=> VH2 = (0,1 + 0,05) . 22,4 = 3,36 (l)
c, Theo pt (1): nCuO = nCu = 0,05 (mol)
Theo pt (2): nFe2O3 = \(\dfrac{1}{2}n_{Fe}=\dfrac{1}{2}.0,05=0,025\left(mol\right)\)
=> m = 0,05 . 80 + 0,025 . 160 = 8 (g)
\(a.CuO+H_2-^{t^o}\rightarrow Cu+H_2O\\ Fe_2O_3+3H_2-^{t^o}\rightarrow2Fe+3H_2O\\ b.m_{Cu}=6-2,8=3,2\left(g\right)\\ n_{Cu}=0,05\left(mol\right);n_{Fe}=0,05\left(mol\right)\\ \Sigma n_{H_2}=n_{Cu}+\dfrac{3}{2}n_{Fe}=0,125\left(mol\right)\\ \Rightarrow V_{H_2}=2,8\left(l\right)\\ c.n_{CuO}=n_{Cu}=0,05\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,025\left(mol\right)\\ m_{hh}=m_{CuO}+m_{Fe_2O_3}=0,05.80+0,025.160=8g\)
a, Ta có:
\(n_{HgO}=\frac{21,7}{217}=0,1\left(mol\right)\)
\(\Rightarrow n_{O2}=n_{HgO}=0,1\left(mol\right)\)
\(\Rightarrow V_{O2}=0,1.22,4=2,24\left(l\right)\)
b, Ta có:
\(n_{HgO}=0,2\left(mol\right)\Rightarrow n_{Hg}=n_{HgO}=0,2\left(mol\right)\)
\(\Rightarrow m_{Hg}=0,2.201=40,2\left(g\right)\)
c,Ta có:
\(n_{Hg}=0,07\left(mol\right)\Rightarrow n_{HgO}=0,07\left(mol\right)\)
\(\Rightarrow m_{HgO}=0,07.217=15,19\left(g\right)\)
câu c sai