$n_{Ca(OH)_2} = \dfrac{185.20\%}{74} = 0,5(mol)$

$n_{CaCO_3} = 0,3(mol)$

TH1 : Ca(OH)₂ dư

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

0,3..............0,3.........0,3........................(mol)

V = 0,3.22,4 = 6,72 lít

$m_{dd} = 0,3.44 + 185 - 30 = 168,2(gam)$

$C\%_{Ca(OH)_2\ dư} = \dfrac{(0,5 - 0,3).74}{168,2}.100\% = 8,8\%$

TH2 : Có tạo muối axit

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

0,3..............0,3.........0,3........................(mol)

Ca(OH)₂ + 2CO₂ → Ca(HCO₃)₂

0,2.............0,4...............0,2.........................(mol)

V = (0,3 + 0,4).22,4 = 15,68 lít

$m_{dd} = 0,7.44 + 185 - 30 = 185,8(gam)$

$C\%_{Ca(HCO_3)_2} = \dfrac{0,2.162}{185,8}.100\% = 17,4\%$

Theo gt ta có: $n_{Ca(OH)_2}=\frac{92}{185}(mol)$

+, Xét trường hợp 1: Chỉ tạo muối $CaCO_3$

$\Rightarrow n_{CO_2}=n_{CaCO_3}=0,3(mol)\Rightarrow V=6,72(l)$

Bảo toàn khối lượng ta có: $m_{dd}=167,2(g)$

$\Rightarrow \%C_{Ca(OH)_2}=8,73\%$

+, Xét trường hợp 2: Tạo 2 muối

Bảo toàn Ca ta có: $n_{Ca(HCO_3)_2}=\frac{73}{370}(mol)$

Bảo toàn C ta có: $n_{CO_2}=\frac{257}{370}(mol)\Rightarrow V_{CO_2}=15,56(l)$

$\Rightarrow \%m_{Ca(HCO_3)_2}=17,3\%$

\(a)n_{H_2SO_4}=\dfrac{58,8.20}{100.98}=0,12mol\\ n_{BaCl_2}=\dfrac{200.5,2}{100.208}=0,05mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{0,12}{1}>\dfrac{0,05}{2}\Rightarrow H_2SO_4.dư\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

0,05         0,05             0,05           0,1

\(m_{BaSO_4}=0,05.233=11,65g\\ b)m_{dd}=58,8+200-11,65=247,15g\\ C_{\%HCl}=\dfrac{0,1.36,5}{247,15}\cdot100=1,48\%\\ C_{\%H_2SO_4,dư}=\dfrac{\left(0,12-0,05\right).98}{247,15}\cdot100=2,78\%\)

\(n_{BaSO_4}=\dfrac{58.25}{233}=0.25\left(mol\right)\)

\(BaCl_2+SO_3+H_2O\rightarrow BaSO_4+2HCl\)

\(0.25........0.25.......................0.25........0.5\)

\(V_{SO_3}=0.25\cdot22.4=5.6\left(l\right)\)

\(m_{dd_{BaCl_2}}=\dfrac{0.25\cdot208}{20\%}=260\left(g\right)\)

\(m_{dd}=m_{SO_3}+m_{dd_{BaCl_2}}-m_{BaSO_4}=0.25\cdot80+260-58.25=221.75\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.5\cdot36.5}{221.75}\cdot100\%=8.2\%\)

chỉ mình câu b,c nữa ạ

a, \(H_2SO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+2HCl\)

b, Ta có: \(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{22,8}{96}=0,2375\left(mol\right)\)

\(m_{BaCl_2}=400.5,2\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2375}{1}>\dfrac{0,1}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,1\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2375-0,1=0,1375\left(mol\right)\)

Ta có: m _{dd sau pư} = 114 + 400 - 23,3 = 490,7 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1375.98}{490,7}.100\%\approx2,75\%\\C\%_{HCl}=\dfrac{0,2.36,5}{490,7}.100\%\approx1,49\%\end{matrix}\right.\)

a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

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

b, \(n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c, \(n_{NaOH}=2n_{CuCl_2}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%\)

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

a) Ta có: \(n_{CuSO_4}=\dfrac{320\cdot20\%}{160}=0,4\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)

b) Theo PTHH: \(n_{NaOH}=2n_{CuSO_4}=0,8mol\) \(\Rightarrow m_{ddNaOH}=\dfrac{0,8\cdot40}{10\%}=320\left(g\right)\)

c) Theo PTHH: \(n_{Na_2SO_4}=n_{CuSO_4}=0,4mol\) \(\Rightarrow m_{Na_2SO_4}=0,4\cdot142=56,8\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=600,8\left(g\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{56,8}{600,8}\cdot100\%\approx9,45\%\)