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\(\)1.
\(n_{O_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(0.8........0.6..........0.4\)
\(m_{Al}=0.8\cdot27=21.6\left(g\right)\)
\(m_{Al_2O_3}=0.4\cdot102=40.8\left(g\right)\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(1.2..................................................0.6\)
\(m_{KMnO_4}=1.2\cdot158=189.6\left(g\right)\)
2.
\(n_{O_2}=\dfrac{28}{22.4}=1.25\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
\(1......1.25........0.5\)
\(m_P=1\cdot31=31\left(g\right)\)
\(m_{P_2O_5}=0.5\cdot142=71\left(g\right)\)
\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)
\(\dfrac{5}{6}................1.25\)
\(m_{KClO_3}=\dfrac{5}{6}\cdot122.5=102.083\left(g\right)\)
Theo ĐLBTKL: mAl2O3 = mAl + mO2
=> mAl2O3 = 10,8 + 9,6 = 20,4 (g)
=> \(\%Al_2O_3=\dfrac{20,4}{21,2}.100\%=96,226\%\)
Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
_____0,4____0,3___0,2 (mol)
b, \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
c, \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)
Bạn tham khảo nhé!
a.\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,6 0,45 0,3 ( mol )
\(m_{Al}=0,6.27=16,2g\)
\(V_{O_2}=0,45.22,4=10,08l\)
\(V_{kk}=10,08.5=50,4l\)
b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,3 0,45 ( mol )
\(m_{KClO_3}=0,3.122,5=36,75g\)
c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,3 0,45 ( mol )
\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)
\(m_{KClO_3}=0,4.122,5=49g\)
