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\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{O_2} = \dfrac{22,4}{22,4} = 1(mol)\\ n_{KMnO_4} = 2n_{O_2} = 2(mol)\\ \Rightarrow H = \dfrac{2.158}{200}.100\% = 158\%>100\%\)
(Sai đề)
2KMnO4-to>K2MnO4+MnO2+O2
0,2---------------------------------------0,1 mol
n O2=\(\dfrac{2,24}{22,4}\)=0,1 mol
=>m KMnO4 tt =0,2.158=31,6g
=>H =\(\dfrac{31,6}{39,5}.100\)=80%
Ta có: \(n_{KMnO_4}=\dfrac{39,5}{158}=0,25\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{O_2\left(LT\right)}=\dfrac{1}{2}n_{KMnO_4}=0,125\left(mol\right)\)
Mà: \(n_{O_2\left(TT\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow H\%=\dfrac{0,1}{0,125}.100\%=80\%\)
Bạn tham khảo nhé!
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%
\(\rightarrow V_{O_2}=0,0115.22,4.80\%=0,20608l\)
nK2MnO4 = 14,775/197 = 0,075 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
Mol: 0,15 <--- 0,075
nKMnO4 (ban đầu) = 31,6/158 = 0,2 (mol)
H = 0,15/0,2 = 75%
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)
2K2MnO4-toK2MnO4+MnO2+O2
nKMnO4=15,8/158=0,1 mol
->nK2MnO4=nMnO2=0,05 mol
mcr=0,05.197+0,05.87=14,2g
H,=14,2/14,5=97,7%
\(n_{KMnO_4}=\dfrac{15,8}{158}=0,1mol\)
Gọi \(n_{KMnO_4}=x\)
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
x 1/2 x 1/2 x ( mol )
Ta có:
\(158\left(0,1-x\right)+\dfrac{1}{2}x\left(197+87\right)=14,52\)
\(\Leftrightarrow x=0,08mol\)
\(H=\dfrac{0,08}{0,1}.100=80\%\)
a) $n_{O_2} = 0,15(mol)$
\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
0,3 0,15 0,15 0,15 (mol)
$H = \dfrac{0,15.158}{63,2}.100\% = 37,5\%$
b)
$m_B = 63,2 - 0,15.32 = 58,4(gam)$
$\%m_{K_2MnO_4} = \dfrac{0,15.197}{58,4}.100\% = 50,59\%$
$\%m_{MnO_2} = \dfrac{0,15.87}{58,4}.100\% = 22,35\%$
$\%m_{KMnO_4\ dư} = 100\% -50,59\% -22,35\% = 27,06\%$
nO2 = 48/32 = 1,5 (mol)
PTHH: KMnO4 -t°-> K2MnO4 + MnO2 + O2
3. 1,5
H = 3/4 = 75%