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$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$
$a\bigg)$
$n_{KMnO_4}=\frac{15,8}{158}=0,1(mol)$
Chất rắn sau p/ứ là $K_2MnO_4,MnO_2$
Theo PT: $n_{K_2MnO_4}=n_{MnO_2}=0,05(mol)$
$\to m_{\rm chất\, rắn}=0,05.197+0,05.87=14,2(g)$
$b\bigg)$
Vì $H=80\%\to n_{KMnO_4(p/ứ)}=0,1.80\%=0,08(mol)$
$\to n_{KMnO_4(dư)}=0,02(mol)$
Chất rắn sau p/ứ là $KMnO_4(dư):0,02;K_2MnO_4:0,04;MnO_2:0,04$
$\to m_{\rm chất\, rắn}=0,02.158+0,04.197+0,04.87=14,52(g)$
$c\bigg)$
Bảo toàn KL có:
$m_{O_2}=m_{KMnO_4}-m_{CR}$
$\to m_{O_2}=15,8-14,68=1,12(g)\to n_{O_2}=0,035(mol)$
Theo PT: $n_{KMnO_4(p/ứ)}=2n_{O_2}=0,07(mol)$
$\to H=\dfrac{0,07}{0,1}.100\%=70\%$
Gọi n KMnO4 = a
n KClO3 = b ( mol )
--> 158a + 122,5 b = 43,3
PTHH :
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
0,9b 1,35b
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,9a 0,45a
\(\%Mn=\dfrac{55a}{43,3-32\left(0,45a+1,35b\right)}=24,103\%\)
\(\rightarrow a=0,15\)
\(b=0,16\)
\(m_{KMnO_4}=0,15.158=23,7\left(g\right)\)
\(m_{KClO_3}=0,16.122,5=19,6\left(g\right)\)
\(a)n_{KMnO_4} = a; n_{KClO_3} = b\Rightarrow 158a + 122,5b = 99,95(1)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{O_2} = 0,5a +1,5b = \dfrac{14,56}{22,4}=0,65(2)\\ (1)(2)\Rightarrow a = 0,4 ; b = 0,3\\ \%m_{KMnO_4} = \dfrac{0,4.158}{99,95}.100\% = 63,23\%\\ \%m_{KClO_3} = 100\%-63,23\% = 36,77\%\)
\(n_{K_2MnO_4} = n_{MnO_2} = 0,5a = 0,2(mol)\\ n_{KClO_3} = b = 0,3(mol)\\ m_{hh\ sau\ pư} = 99,95 - 0,65.32 = 79,15(gam)\\ \%m_{K_2MnO_4} = \dfrac{0,2.197}{79,15}.100\% = 49,78\%\\ \%m_{MnO_2} = \dfrac{0,2.87}{79,15},100\% = 21,98\%\\ \%m_{KCl} = 28,24\%\)
ko khó lém bn ơi có câu nèo easy hơm ko
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(\dfrac{x}{158}.........\dfrac{x}{158}........\dfrac{x}{158}\)
\(2Cu+O_2\underrightarrow{t^0}2CuO\)
\(\dfrac{y}{64}...........\dfrac{y}{64}\)
\(m_A=m_B\)
\(\Rightarrow x+y=\dfrac{x}{158}\cdot197+\dfrac{x}{158}\cdot87+\dfrac{80y}{64}\)
\(\Rightarrow x+y=\dfrac{142x}{79}+1.25y\)
\(\Rightarrow0.25y=-\dfrac{63}{79}x\)
\(\Rightarrow\dfrac{x}{y}=-\dfrac{79}{252}\)
c2
a/ 2KMnO4(x)to→K2MnO4(0,5x)+MnO2(0,5x)+O2(0,5x)
Gọi số mol của KMnO4 tham gia phản ứng là x.
⇒mKMnO4=158x(g)
⇒mK2MnO4=0,5x.197=98,5x(g)
⇒mMnO2=0,5x.87=43,5x(g)
⇒22,12−158x+98,5x+43,5x=21,26
⇔x=0,05375(mol)
⇒VO2=0,05375.0,5.22,4=0,602(l)
b/ mKMnO4(pứ)=0,05375.158=8,4925(g)
⇒%KMnO4=8,4925\22,12.100%=38,39%
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{15,8}{158}=0,1\left(mol\right)\)
a. Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}0,1=0,05\left(mol\right)\)
\(\Rightarrow V_{O_2}=n_{O_2}.22,4=0,05.22,4=1,12\left(l\right)\)
b. PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Ta có: \(\dfrac{1}{n_{O_2}}=\dfrac{1}{0,05}\)
\(\dfrac{1}{n_{Fe}}=\dfrac{1}{0,1}\)
\(\Rightarrow\dfrac{1}{n_{O_2}}>\dfrac{1}{n_{Fe}}\)
Vậy Fe dư
Theo PTHH: \(n_{Fe_3O_4}=\dfrac{0,1.1}{3}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=n_{Fe_3O_4}.M_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,73g\)
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:
\(158\left(0,1-x\right)+\dfrac{1}{2}x\left(197+87\right)=14,52\)
\(\Leftrightarrow x=0,08mol\)
\(H=\dfrac{0,08}{0,1}.100=80\%\)