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\(a.CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\\b. n_{CuSO_4}=n_{Na_2SO_4}=0,2\left(mol\right)\\ m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\\ c.BTNT\left(Cu\right):n_{CuO}=n_{CuSO_4}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)
a)
$FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$n_{FeCl_3} = 0,3.2 = 0,6(mol)$
Theo PTHH : $n_{Fe_2O_3} = \dfrac{1}{2}n_{FeCl_3} = 0,3(mol)$
$\Rightarrow m_{Fe_2O_3} = 0,3.160 = 48(gam)$
b) Sau phản ứng, $V_{dd} = 0,3 + 0,3 = 0,6(lít)$
$n_{NaCl} = 3n_{FeCl_3} = 1,8(mol) \Rightarrow C_{M_{NaCl}} = \dfrac{1,8}{0,6} = 3M$
Bài 7 :
200ml = 0,2l
\(n_{CuCl2}=2.0,2=0,4\left(mol\right)\)
Pt : \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl|\)
1 2 1 2
0,4 0,8 0,4 0,8
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O|\)
1 1 1
0,4 0,4
a) \(n_{CuO}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
⇒ \(m_{CuO}=0,4.40=32\left(g\right)\)
b) \(n_{NaCl}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
⇒ \(m_{NaCl}=0,8.58,5=46,8\left(g\right)\)
\(m_{ddCuCl2}=1,35.200=270\left(g\right)\)
\(m_{ddspu}=270+100=370\left(g\right)\)
\(C_{NaCl}=\dfrac{46,8.100}{370}=12,65\)0/0
Chúc bạn học tốt
a, \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, \(n_{MgCl_2}=0,2.0,25=0,05\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,05.40=2\left(g\right)\)
c, \(n_{NaOH}=2n_{MgCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{0,1.40}{15\%}=\dfrac{80}{3}\left(g\right)\)
\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,095 0,19 0,095 0,095
\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)
0,095 0,095
\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)
\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)
PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)
a)\(n_{CuSO_4}=0,4.0,5=0,2\left(mol\right)\)
\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
Mol: 0,2 0,4 0,2
⇒ \(m_{Cu\left(OH\right)_2}=0,2.98=19,6\left(g\right)\)
b)\(C_{M\left(ddNaOH\right)}=\dfrac{0,4}{0,3}=1,3\left(M\right)\)
c)\(PTHH:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
Mol: 0,2 0,2
=> mCuO = 0,2.80 = 16 (g)