Đáp án:

 m=400 gam.

Giải thích các bước giải:

Phản ứng xảy ra:

\(2NaHC{O_3}\xrightarrow{{{t^o}}}N{a_2}C{O_3} + C{O_2} + {H_2}O\)

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)

Ta có: 

\({n_{CaC{O_3}}} = \frac{{20}}{{100}} = 0,2{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)

\( \to {n_{NaHC{O_3}}} = 2{n_{C{O_2}}} = 0,4{\text{ mol}}\)

\( \to {m_{NaHC{O_3}}} = 0,4.84 = 33,6{\text{ gam}}\)

\( \to m = \frac{{33,6}}{{8,4\% }} = 400{\text{ gam}} \to {{\text{m}}_{{H_2}O}} = 366,4{\text{ gam}}\)

Khi nung: \({n_{C{O_2}}} = {n_{{H_2}O}} = 0,2{\text{ mol}}\)

\({m_{{H_2}O{\text{ trong B}}}} = 366,4 + 0,2.18 = 370{\text{ gam}}\)

BTKL: 

\({m_{dd{\text{ tăng}}}} = a = {m_{C{O_2}}} + {m_{{H_2}O}} - {m_{CaC{O_3}}}\)

\( = 370 + 0,2.44 - 20 = 358,8{\text{ gam}}\)

$n_{CaCO_3}=\dfrac{20}{100}=0,2 mol$

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

$\Rightarrow n_{CO_2}=0,2 mol$

$2NaHCO_3\buildrel{{t^o}}\over\to Na_2CO_3+CO_2+H_2O$

$\Rightarrow n_{NaHCO_3}=0,2.2=0,4 mol$

$n_{H_2O(\text{sinh ra})}=0,2 mol$

$m_{dd NaHCO_3}=0,4.84:8,4\%=400g$

$m_{H_2O(dd)}=400-0,4.84=366,4g$

Vậy B chứa:

$CO_2: 0,2.44=8,8g$

$H_2O: 366,4+0,2.18=370g$

$\Rightarrow a=m_{CO_2}+m_{H_2O}-m_{CaCO_3}=8,8+370-20=358,8g$