a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$
b) Theo PTHH : $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{64}{160} = 0,4(mol)$

$m_{Cu(OH)_2} = 0,4.98 = 39,2(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,8(mol)$
$C_{M_{NaOH}} = \dfrac{0,8}{0,4} = 2M$

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)

a)\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

     0,1             0,2        0,1         0,1         0,1

b)\(m_{HCl}=0,1\cdot36,5=3,65\left(g\right)\)

   \(a\%=\dfrac{3,65}{100}\cdot100\%=3,65\%\)

c)\(m_{CaCO_3}=0,1\cdot100=10\left(G\right)\)

  \(\Rightarrow\%m_{CaCO_3}=\dfrac{10}{16}\cdot100\%=62,5\%\)

  \(\Rightarrow\%m_{CaCl_2}=100\%-62,5\%=37,5\%\)

d)\(m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\)

   \(m_{H_2O}=0,1\cdot18=1,8\left(g\right)\)

   \(m_{ddsau}=10+100-0,1\cdot44-1,8=103,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{11,1}{103,8}\cdot100\%=10,7\%\)

\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

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

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

b, \(m_{CuSO_4}=250.16\%=40\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{40}{160}=0,25\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,25\left(mol\right)\)

\(\Rightarrow a=m_{CuO}=0,25.80=20\left(g\right)\)

c, Ta có: m _{dd sau pư} = m _{dd NaOH} + m _{dd CuSO4} - m_Cu(OH)2 = 200 + 250 - 0,25.98 = 425,5 (g)

\(a)2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ Cu\left(OH\right)_2\xrightarrow[]{t^0}CuO+H_2O\\ b)n_{CuSO_4}=\dfrac{250.16}{100.160}=0,25mol\\ n_{CuSO_4}=n_{Cu\left(OH\right)_2}=n_{CuO}=0,25mol\\ a=m_{CuO}=0,25.80=20g\\ c)m_{dd}=200+250-0,25.98=425,5g\)

\(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)

\(PTHH:CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

bđ:           0,3           0,5                

pứ:          0,25          0,5             0,5             0,25

[    ]:        0,05          0                 0,5             0,25

\(PTHH:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

(mol)         0,25            0,25

\(a.m_C=80.0,25=20\left(g\right)\)

\(b.m_{NaCl}=58,5.0,5=29,25\left(g\right)\\ m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\\ m_{CuCl_2\left(du\right)}=135.0,05=6,75\left(g\right)\)

\(c.m_{ddspu}=100+200-24,5=275,5\left(g\right)\\ C\%_{ddCuCl_2\left(du\right)}=\dfrac{135.0,05}{275,5}.100=2,45\left(\%\right)\\ C\%_{ddNaCl}=\dfrac{0,5.58,5}{275,25}.100=10,62\left(\%\right)\)

a/

\(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,15 0,3 (mol)

\(m_{NaOH}=0,3.40=12\left(g\right)\)

\(m_A=90,7+9,3=100\left(g\right)\)

\(C\%_{NaOH}=\dfrac{12}{100}.100\%=12\%\)

b/

m\(_{FeSO_4}=\dfrac{16.200}{100}=32\left(g\right)\)

\(\rightarrow m_{FeSO_4}=\dfrac{32}{152}=\dfrac{4}{19}\left(mol\right)\)

\(2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\downarrow\)

bđ: 0,3 \(\dfrac{4}{19}\) 0 0 (mol)

pư: 0,3 0,15 0,15 0,15 (mol)

dư: 0 \(\dfrac{23}{380}\) (mol)

\(m_{Fe\left(OH\right)_2}=0,15.90=13,5\left(g\right)\)

\(m_C=100+200-13,5=286,5\left(g\right)\)

\(m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)

\(\rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{286,5}.100\%\approx7,4\%\)

\(m_{FeSO_4\left(dư\right)}=\dfrac{23}{380}.152=9,2\left(g\right)\)

\(\rightarrow C\%_{FeSO_4\left(dư\right)}=\dfrac{9,2}{286,5}.100\%\approx3,2\%\)

a.

\(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)

b.

\(n_{BaCO_3}=n_{Na_2CO_3}=0,2.1=0,2\left(mol\right)\\ m_{kt}=197.0,2=39,4\left(g\right)\)

c.

\(n_{Ba\left(OH\right)_2}=n_{Na_2CO_3}=0,2\left(mol\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,2.171.100\%}{200}=17,1\%\)

\(a,Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{Mg}=\dfrac{12}{24}=0,5mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,5mol\\ V_{H_2}=0,5.24,79=12,395l\\ c,n_{HCl}=0,5.2=1mol\\ C_{M_{HCl}}=\dfrac{1}{0,1}=10M\\ d,m_{MgCl_2}=0,5.95=47,5g\)

`#3107.101107`

`a,`

PTHH: \(\text{Mg}+\text{2HCl}\rightarrow\text{ MgCl}_2+\text{H}_2\)

`b,` 

n của Mg trong pứ là:

\(\text{n}_{\text{Mg}}=\dfrac{\text{m}_{\text{Mg}}}{\text{M}_{\text{Mg}}}=\dfrac{12}{24}=\dfrac{1}{2}=0,5\left(\text{mol}\right)\)

Theo pt: `1` mol Mg thu được `1` mol H₂

`=>`\(\text{n}_{\text{H}_2}=0,5\left(\text{mol}\right)\)

Thể tích của H₂ thu được sau pứ là:

\(\text{V}_{\text{H}_2}=\text{n}_{\text{H}_2}\cdot24,79=0,5\cdot24,79=12,395\left(l\right)\)

c.

Đổi `100` ml = `0,1` l

Theo PT: `1` mol Mg tác dụng với `2` mol HCl được pứ trên

`=>`\(\text{n}_{\text{HCl}}=2\cdot0,5=1\left(\text{mol}\right)\)

Nồng độ `%` của HCl là:

\(\text{C}_{\text{M}}=\dfrac{\text{n}_{HCl}}{\text{V}_{\text{HCl}}}=\dfrac{1}{0,1}=10\left(\text{M}\right)\)

d.

Theo PT: `1` mol Mg thu được `1` mol MgCl₂

`=>`\(\text{n}_{\text{MgCl}_2}=0,5\text{ (mol)}\)

m của MgCl₂ thu được sau pứ là:

\(\text{m}_{\text{MgCl}_2}=\text{n}_{\text{MgCl}_2}\cdot\text{M}_{\text{MgCl}_2}=0,5\cdot95=47,5\left(g\right).\)