$2R(HCO_3)_n\buildrel{{t^o}}\over\to R_2(CO_3)_n+ nCO_2+nH_2O$

(Nếu muối cacbonat bị nhiệt phân tiếp) $R_2(CO_3)_n\buildrel{{t^o}}\over\to R_2O_n+ nCO_2$

$\Rightarrow n_{CO_2}\ge n_{H_2O}$

$n_{CaCO_3}=0,04(mol)<n_{Ca(OH)_2}$

- TH1: dư kiềm 

$Ca(OH)_2+CO_2\to CaCO_3+H_2O$

$\Rightarrow n_{CO_2}=0,04(mol)$

$m_{CO_2}=0,04.44=1,76g$

$\Rightarrow n_{H_2O}=\dfrac{5,3-1,76}{18}=0,196>0,04(mol)$ (loại)

- TH2: tạo 2 muối

$Ca(OH)_2+CO_2\to CaCO_3+H_2O$

$Ca(OH)_2+2CO_2\to Ca(HCO_3)_2$

$n_{Ca(OH)_2}=n_{CaCO_3}+n_{Ca(HCO_3)_2}$

$\Rightarrow n_{Ca(HCO_3)_2}=0,03(mol)$

$\Rightarrow n_{CO_2}=n_{CaCO_3}+2n_{Ca(HCO_3)_2}=0,1(mol)$

$m_{CO_2}=0,1.44=4,4g$

$\Rightarrow n_{H_2O}=\dfrac{5,3-4,4}{18}=0,05(mol)$ (TM)

Ta có $n_{R(HCO_3)_n}=\dfrac{2n_{H_2O}}{n}=\dfrac{0,1}{n}(mol)$

$\Rightarrow M=\dfrac{12,95n}{0,1}=129,5n=R+61n$

$\Leftrightarrow R=68,5n$

$n=2\Rightarrow R=137(Ba)$

$n_{BaO}=n_{Ba(HCO_3)_2}=0,05(mol)$

$D_{H_2SO_4}=1,2g/ml, V=100ml$

$\Rightarrow m_{dd H_2SO_4}=100.1,2=120g$

$n_{H_2SO_4}=0,1.0,2=0,02(mol)$

$BaO+H_2SO_4\to BaSO_4+H_2O$

$\Rightarrow n_{BaSO_4}=0,02(mol)$

$BaO+H_2O\to Ba(OH)_2$

$n_{Ba(OH)_2}=0,05-0,02=0,03(mol)$

$\Rightarrow C\%_{Ba(OH)_2}=\dfrac{0,03.171.100}{0,05.153+120-0,02.233}=4,17\%$ 

Đáp án:

 đây