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nKMnO4 = \(\frac{m}{M}=\frac{15,8}{158}=0,1\left(mol\right)\)
nO2 = \(\frac{V\left(\text{đ}kc\right)}{22,4}=\frac{0,784}{22,4}=0,035\left(mol\right)\)
PTPU : 2KMnO4 \(\rightarrow\) O2\(\uparrow\) + K2MnO4 + MnO2
PU : 0,1 _______ 0,035___ ________________ (mol)
\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{O_2} = \dfrac{22,4}{22,4} = 1(mol)\\ n_{KMnO_4} = 2n_{O_2} = 2(mol)\\ \Rightarrow H = \dfrac{2.158}{200}.100\% = 158\%>100\%\)
(Sai đề)
a. PTHH: \(KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
b. \(H=100\%\)
\(n_{KMnO_4}=\frac{3,6}{158}=0,023mol\)
Theo phương trình \(n_{O_2}=0,5n_{KMnO_4}=0,046mol\)
\(\rightarrow V_{O_2}=0,0115.22,4.100\%=0,2576l\)
c. H = 80%
\(\rightarrow V_{O_2}=0,0115.22,4.80\%=0,20608l\)
\(n_{KMnO_4}=\dfrac{118,5}{158}=0,75\left(mol\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,75 0,375
=> \(V_{O_2}=0,375.22,4=8,4\left(l\right)\\
V_{kk}=8,4.5=42\left(l\right)\)
a, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
b, \(n_{KCl}=\dfrac{0,745}{74,5}=0,01\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,015\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,015.24,79=0,37185\left(l\right)\)
\(m_{O_2}=0,015.32=0,48\left(g\right)\)
c, \(n_{KClO_3\left(pư\right)}=n_{KCl}=0,01\left(mol\right)\)
\(\Rightarrow m_{KClO_3\left(pư\right)}=0,01.122,5=1,225\left(g\right)\)
\(\Rightarrow H=\dfrac{1,225}{2,5}.100\%=49\%\)
\(n_{KMnO_4}=\dfrac{79}{158}=0,5mol\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,5 0,25
\(H=80\%\Rightarrow n_{O_2}=0,25\cdot80\%=0,2mol\)
\(\Rightarrow V=0,2\cdot22,4=4,48l\)