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\(\begin{array}{l} a,\\ PTHH:\\ AlCl_3+3KOH\to Al(OH)_3\downarrow+3KCl\ (1)\\ 2Al(OH)_3\xrightarrow{t^o} Al_2O_3+3H_2O\ (2)\\ b,\\ n_{KOH}=\dfrac{3,36}{56}=0,06\ (mol)\\ Theo\ pt\ (1):\ n_{AlCl_3}=\dfrac{1}{3}n_{KOH}=0,02\ (mol)\\ \Rightarrow m_{AlCl_3}=0,02\times 133,5=2,67\ (g)\\ c,\\ Theo\ pt\ (1):\ n_{Al(OH)_3}=\dfrac{1}{3}n_{KOH}=0,02\ (mol)\\ Theo\ pt\ (2):\ n_{Al_2O_3}=\dfrac{1}{2}n_{Al(OH)_3}=0,01\ (mol)\\ \Rightarrow m_{Al_2O_3}=0,01\times 102=1,02\ (g)\end{array}\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
2NaOH + CuSO4 -> Cu(OH)2 + Na2SO4
x 0,5x
2KOH + CuSO4 -> Cu(OH)2 + K2SO4
y 0,5y
Cu(OH)2 -> CuO + H2O
0,125 0,125
nCuO= 10/80 = 0,125 mol
đặt x=nNaOH , y=nKOH mol
theo đề bài , ta có hệ pt
40x + 56y = 11,6 -> x=0,15
0,5x + 0,5y = 0,125 y=0,1
mNaOH= 0,15*40=6 g
mKOH= 5,6 g
Chúc bạn thành công nha =))))