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a)
Gọi $n_{KMnO_4} = a(mol) \Rightarrow n_{KClO_3} = 2a(mol)$
Suy ra :
$158a + 122,5.2a = 40,3 \Rightarrow a = 0,1(mol)$
$m_{KMnO_4} = 0,1.158 = 15,8(gam)$
$m_{KClO_3} = 0,1.122,5 = 12,25(gam)$
b)
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
Theo PTHH :
$n_{O_2} = \dfrac{1}{2}n_{KMnO_4} + \dfrac{3}{2}n_{KClO_3} = 0,35(mol)$
$m_{O_2} = 0,35.32 = 11,2(gam)$
a)
$2Mg + O_2 \xrightarrow{t^o} 2MgO$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$2Zn + O_2 \xrightarrow{t^o} 2ZnO$
b)
Bảo toàn khối lượng :
$m_{O_2\ pư} = 8,4 - 5,2 = 3,2(gam)$
$n_{O_2\ pư} = \dfrac{3,2}{32} = 0,1(mol)$
$V_{O_2\ pư} = 0,1.22,4 = 2,24(lít)$
\(a)n_{KMnO_4} = a; n_{KClO_3} = b\Rightarrow 158a + 122,5b = 99,95(1)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{O_2} = 0,5a +1,5b = \dfrac{14,56}{22,4}=0,65(2)\\ (1)(2)\Rightarrow a = 0,4 ; b = 0,3\\ \%m_{KMnO_4} = \dfrac{0,4.158}{99,95}.100\% = 63,23\%\\ \%m_{KClO_3} = 100\%-63,23\% = 36,77\%\)
\(n_{K_2MnO_4} = n_{MnO_2} = 0,5a = 0,2(mol)\\ n_{KClO_3} = b = 0,3(mol)\\ m_{hh\ sau\ pư} = 99,95 - 0,65.32 = 79,15(gam)\\ \%m_{K_2MnO_4} = \dfrac{0,2.197}{79,15}.100\% = 49,78\%\\ \%m_{MnO_2} = \dfrac{0,2.87}{79,15},100\% = 21,98\%\\ \%m_{KCl} = 28,24\%\)
ko khó lém bn ơi có câu nèo easy hơm ko
\(BTKL:\)
\(m_A+m_{O_2}=m_B\)
\(\Rightarrow m_{O_2}=m_B-m_A=32-22.4=9.6\left(g\right)\)
\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\) (1)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (2)
a) Gọi số mol của Mg là a (mol) \(\Rightarrow n_{Al}=\dfrac{2}{3}a\left(mol\right)\)
\(\Rightarrow24a+27\cdot\dfrac{2}{3}a=6,3\) \(\Rightarrow a=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{MgO}=0,15\left(mol\right)\\n_{Al_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgO}=0,15\cdot40=6\left(g\right)\\m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\end{matrix}\right.\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,075\left(mol\right)\\n_{O_2\left(2\right)}=0,075\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
2 kim loại gồm Fe, Cu
\(n_{Al}=a;n_{Fe\left(pư\right)}=b;n_{Fe\left(dư\right)}=c\\ 27a+56\left(b+c\right)=8,3\\ n_{Cu}=0,2.1,05=0,21=1,5a+b\\ m_X=56c+64.0,21=15,68\\ a=0,1;b=0,06;c=0,04\\ \%m_{Al}=\dfrac{27a}{8,3}.100\%=32,53\%\\ \%m_{Fe}=67,47\%\)
a, \(2Cu+O_2\underrightarrow{t^o}2CuO\)
\(2Mg+O_2\underrightarrow{t^o}2MgO\)
b, \(\dfrac{m_{CuO}}{m_{MgO}}=\dfrac{2}{1}\Rightarrow\dfrac{n_{CuO}}{n_{MgO}}=\dfrac{2}{1}:\dfrac{80}{40}=1\)
⇒ nCuO = nMgO (1)
Có: m chất rắn tăng = mO2 = 32 (g)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{CuO}+\dfrac{1}{2}n_{MgO}=\dfrac{32}{32}=1\left(mol\right)\left(2\right)\)
Từ (1) và (2) ⇒ nCuO = nMgO = 1 (mol)
⇒ mCuO = 1.80 = 80 (g)
mMgO = 1.40 = 40 (g)