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PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)
a+b) Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,2\left(mol\right)=n_{KOH}\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2\cdot36,5}{300}\cdot100\%\approx2,43\%\\C_{M_{KOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
c) PTHH: \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Theo các PTHH: \(n_{CuO\left(lý.thuyết\right)}=n_{Cu\left(OH\right)_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow n_{CuO}=0,1\cdot95\%=0,095\left(mol\right)\) \(\Rightarrow m_{CuO}=0,095\cdot80=7,6\left(g\right)\)
1)
a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)
PTHH: Al2O3 + 6HCl → 2AlCl3 + 3H2O
Mol: 0,15 0,9 0,3
\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)
b, mdd sau pứ = 15,3 + 164,25 = 179,55 (g)
c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)
2)
a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: Al2O3 + 6HCl → 2AlCl3 + 3H2O
Mol: 0,05 0,3 0,1
\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
b, mdd sau pứ = 5,1 + 54,75 = 59,85 (g)
\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)
a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$
b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)
Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$
$n_{CaO} = 0,2(mol)$
$n_{HCl} = 0,2V(mol) ; n_{HNO_3} = 0,2V(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
$CaO + 2HNO_3 \to Ca(NO_3)_2 + H_2O$
Theo PTHH :
$n_{axit} = 0,2V + 0,2V = 2n_{CaO} = 0,4 \Rightarrow V = 1(lít)$
Theo PTHH :
$n_{H_2O} = \dfrac{1}{2}n_{axit} = 0,2(mol)$
Bảo toàn khối lượng :
$m_{muối} = 11,2 + 0,2.63 + 0,2.36,5 - 0,2.18 = 27,5(gam)$
\(m_{ct}=\dfrac{20.98}{100}=19,6\left(g\right)\)
\(n_{H2SO4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)
1 1 1 1
0,2 0,2 0,2
a) \(n_{CuO}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CuO}=0,2.80=16\left(g\right)\)
b) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)
c) \(m_{ddspu}=16+98=114\left(g\right)\)
\(C_{CuSO4}=\dfrac{32.100}{114}=28,7\)0/0
Chúc bạn học tốt
a: \(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(V_{H_2}=0,2\cdot22,4=4,48\left(lít\right)\)
b: \(\dfrac{n_{HCl}}{V_{HCl}}=2\)
=>\(\dfrac{0.4}{V_{HCl}}=2\)
=>\(V_{HCl}=\dfrac{0.4}{2}=0.2\left(lít\right)\)
c: \(C_M=\dfrac{n}{V}=\dfrac{0.2}{0.2}=1\)