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\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\\ n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ \Rightarrow n_{KOH}=2n_{CuCl_2}=0,4\left(mol\right)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4}{0,2}=2M\\ b,PTHH:Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)
\(n_{CuCl_2}=2.0,2=0,4(mol)\\ n_{NaOH}=2.0,2=0,4(mol)\\ a,CuCl_2+2NaOH\to Cu(OH)_2+2NaCl\\ Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ b,\dfrac{n_{CuCl_2}}{1}>\dfrac{n_{NaOH}}{2}\Rightarrow CuCl_2\text{ dư}\\ \Rightarrow n_{CuO}=0,2(mol)\\ \Rightarrow m_{CuO}=0,2.80=16(g)\\ c,n_{CuCl_2(dư)}=0,4-0,2=0,2(mol)\\n_{NaCl}=0,2(mol)\\ \Rightarrow m_{CuCl_2(dư)}=0,2.135=27(g)\\ m_{NaCl}=0,2.58,5=11,7(g)\)
PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)
a) \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
Lập tỉ lệ : \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)=> Sau phản ứng NaOH dư
\(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\)
Dung dịch nước lọc gồm NaCl (0,4_mol); NaOH dư ( 0,1 mol)
\(Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\)
\(a=m_{CuO}=0,2.80=16\left(g\right)\)
b) \(m_{NaCl}=0,4.58,5=23,4\left(g\right);m_{NaOH}=0,1.40=4\left(g\right)\)
PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2\cdot2=0,4\left(mol\right)\\n_{NaOH}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) \(\Rightarrow\) CuCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,4\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}=n_{CuCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2\cdot80=16\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\\\end{matrix}\right.\)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
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: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
a.CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl
0.15 0.3 0.15 0.3
Cu(OH)2 -> CuO + H2O
0.15 0.15
nNaOH = 0.3 mol
\(CM_{CuCl2}=\dfrac{0.15}{2}=0.075M\)
b.Vdd sau phản ứng = 0.2 + 0.15 = 0.35l
\(CM_{NaCl}=\dfrac{0.3}{0.35}=0.86M\)
c.mCuO = \(0.15\times80=12g\)