a, \(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

b, \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

c, \(n_{\left(CH_3COO\right)_2Cu}=n_{Cu}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Cu}=0,1.182=18,2\left(g\right)\)

`n_{Na_2O}={4,6}/{62}\approx 0,074(mol)`

`Na_2O+H_2O->2NaOH`

`0,074->0,074->0,148(mol)`

`a)\ C_{M\ NaOH}={0,148}/{0,2}=0,74M`

`b)\ C\%_{NaOH}={0,148.40}/{200.1,12}.100\%\approx 2,64\%`

\(n_{CuSO4}=\dfrac{16\%.50}{100\%.160}=0,05\left(mol\right)\)

Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

       0,05-------->0,1---------->0,05--------->0,05

a) \(C\%_{ddNaOH}=\dfrac{0,1.40}{250}.100\%=1,6\%\)

b) \(m_{ddspu}=50+250-0,05.98=295,1\left(g\right)\)

\(C\%_{Na2SO4}=\dfrac{0,05.142}{295,1}.100\%=2,41\%\)

\(n_{CuSO_4}=\dfrac{m_{dd}\cdot C\%}{100\cdot M}=\dfrac{50\cdot16\%}{100\cdot\left(64+32+16\cdot4\right)}=0,05\left(mol\right)\)

\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

                     1                  2                    1                    1

                   0,05              0,1                 0,05               0,05 (mol)

\(a)C\%_{NaOH}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{01\cdot100\cdot\left(23+16+1\right)}{250}=1,6\%\)

\(b)m_{dd-sau-pư}=m_{dd_đ}+m_{ct_đ}-m\downarrow-m\uparrow\)

\(=50+250-\left(0,05\cdot23+32+16\cdot4\right)=294,05\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{0,05\cdot100\cdot\left(23\cdot2+32+16\cdot4\right)}{294,05}\approx2,41\%.\)

nCuSO4​=100%.16016%.50​=0,05(mol)

Pt : ����4+2����→��(��)2+��2��4CuSO4​+2NaOH→Cu(OH)2​+Na2​SO4​

       0,05-------->0,1---------->0,05--------->0,05

a) �%������=0,1.40250.100%=1,6%C%ddNaOH​=2500,1.40​.100%=1,6%

b) ������=50+250−0,05.98=295,1(�)mddspu​=50+250−0,05.98=295,1(g)

�%��2��4=0,05.142295,1.100%=2,41%C%Na2SO4​=295,10,05.142​.100%=2,41%

c1 n NaOH=8:40=0,2 mol

CM DUNG DỊCH =0,2:0,8=0,25 M

Bài 1

\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

Bài 5

\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

Zn + 2HCl --> ZnCl2 + H2

0,5      1              0,5       0,5

nZn=\(\dfrac{32,5}{65}\)=0,5(mol)

mZnCl2=0,5.136=68(g)

VH2=0,5.22,4=11,2(l)

c) CM HCl=\(\dfrac{1}{0,25}=4M\)

Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)

a, \(C_{M_{CH_3COOH}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)

a)

$K_2SO_4 + BaCl_2 \to BaSO_4 + 2KCl$

b)

$n_{K_2SO_4} = 0,2.2 = 0,4(mol)$
$n_{BaCl_2} = 0,3.1 = 0,3(mol)$

Ta thấy : 

$n_{K_2SO_4} : 1 > n_{BaCl_2} : 1$ nên $K_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,3(mol)$
$m_{BaSO_4} = 0,3.233 = 69,9(gam)$

c) $n_{K_2SO_4} = 0,4 - 0,3 = 0,1(mol)$

$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$

$C_{M_{K_2SO_4} } = \dfrac{0,1}{0,5} = 0,2M$
$C_{M_{KCl}} = \dfrac{0,6}{0,5} = 1,2M$