Đáp án:

\(\begin{array}{l}
a)\\
CTHH:N{a_2}C{O_3}\\
b)\\
{C_{{M_{HCl}}}} = 0,28M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{R_2}C{O_3} + 2HCl \to 2RCl + C{O_2} + {H_2}O(1)\\
{n_{C{O_2}}} = \dfrac{{0,112}}{{22,4}} = 0,005mol\\
{n_{{R_2}C{O_3}}} = {n_{C{O_2}}} = 0,005mol\\
 \Rightarrow {M_{{R_2}C{O_3}}} = \frac{{0,53}}{{0,005}} = 106dvC\\
 \Rightarrow 2{M_R} + {M_C} + 3{M_O} = 106\\
 \Rightarrow {M_R} = \dfrac{{106 - 12 - 3 \times 16}}{2} = 23dvC\\
 \Rightarrow R:Natri(Na)\\
CTHH:N{a_2}C{O_3}\\
b)\\
{n_{HCl(1)}} = 2{n_{C{O_2}}} = 0,01mol\\
2HCl + Ba{(OH)_2} \to BaC{l_2} + 2{H_2}O(2)\\
{n_{Ba{{(OH)}_2}}} = 0,15 \times 0,2 = 0,03mol\\
 \Rightarrow {n_{HCl(2)}} = 2{n_{Ba{{(OH)}_2}}} = 0,06mol\\
{n_{HCl}} = {n_{HCl(1)}} + {n_{HCl(2)}} = 0,01 + 0,06 = 0,07mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,07}}{{0,25}} = 0,28M
\end{array}\)

a,

$n_{CO_2}=\dfrac{0,112}{22,4}=0,005(mol)$

Gọi CTHH muối là $R_2CO_3$.

$R_2CO_3+2HCl\to 2RCl+CO_2+H_2O$

$\Rightarrow n_{R_2CO_3}=0,005(mol)$

$M_{R_2CO_3}=\dfrac{0,53}{0,005}=106=2M_R+60$

$\Leftrightarrow M_R=23(Na)$

Vậy muối là $Na_2CO_3$

b,

$n_{HCl\text{pứ}}=0,005.2=0,01(mol)$

$n_{Ba(OH)_2}=0,15.0,2=0,03(mol)$

$Ba(OH)_2+2HCl\to BaCl_2+2H_2O$

$\Rightarrow n_{HCl\text{dư}}=0,03.2=0,06(mol)$

$C_{M_{HCl}}=\dfrac{0,06+0,01}{0,25}=0,28M$