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a) PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,075\left(mol\right)\) \(\Rightarrow V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\)
b) PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow m_{KMnO_4}=0,15\cdot138=23,7\left(g\right)\)
a)
\(n_{KClO_3}=\dfrac{24.5}{122.5}=0.2\left(mol\right)\)
\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)
\(n_{O_2}=\dfrac{3}{2}\cdot0.2=0.3\left(mol\right)\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(0.2...............................................0.1\)
\(n_{KMnO_4\left(bđ\right)}=\dfrac{0.2}{90\%}=\dfrac{2}{9}\left(mol\right)\)
\(m_{KMnO_4}=\dfrac{2}{9}\cdot158=35.11\left(g\right)\)
\(a/n_{Fe}=\dfrac{2,52}{56}=0,045mol\\ 3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ n_{O_2}=\dfrac{0,045.2}{3}=0,03mol\\ V_{O_2}=0,03.22,4=0,672l\\ b/2KClO_3\xrightarrow[]{t^0}2KCl+3O_2\\ n_{KClO_3}=\dfrac{0,03.2}{3}=0,02mol\\ m_{KClO_3}=0,02.122,5=2,45g\)
\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ n_{O_2}=3.0,4=1,2\left(mol\right);n_{CO_2}=0,4.2=0,8\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=22,4.1,2=26,88\left(l\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.26,88=134,4\left(l\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow\left(trắng\right)+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,8\left(mol\right)\\ m_{kết.tủa}=m_{CaCO_3}=100.0,8=80\left(g\right)\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(nO_2=3.0,25=0,75\left(mol\right)\)
\(VO_2=0,75.22,4=16,8\left(l\right)\)
\(nCO_2=2.0,25=0,5\left(mol\right)\)
\(VCO_2=0,5.224=11,2\left(l\right)\)
a, \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(n_{O_2}=3n_{C_2H_4}=1,5\left(mol\right)\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)
b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=168\left(l\right)\)
a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)$
$V_{O_2} = 0,15.22,4 = 3,36(lít)$
b) $2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2$
$n_{KClO_3\ pư} = \dfrac{2}{3}n_{O_2} = 0,1(mol)$
$m_{KClO_3\ pư} = 0,1.122,5 = 12,25(gam)$
$\Rightarrow m_{KClO_3\ đã\ dùng} = 12,25 : (100\% - 10\%) = 13,61(gam)$