Đáp án:

 m=18,9 gam.

Giải thích các bước giải:

 Phản ứng xảy ra:

\(N{a_2}S{O_3} + CaC{l_2}\xrightarrow{{}}CaS{O_3} + 2NaCl\)

Ta có: 

\({m_{CaC{l_2}}} = 111.18\%  = 19,98{\text{ gam}} \to {{\text{n}}_{CaC{l_2}}} = \frac{{19,98}}{{111}} = 0,18{\text{ mol}}\)

Gọi số mol \(N{a_2}S{O_3}\) phản ứng là x.

\( \to {n_{N{a_2}S{O_3}}} = {n_{CaS{O_3}}} = x\)

\( \to {m_{dd{\text{ tăng}}}} = {m_{N{a_2}S{O_3}}} - {m_{CaS{O_3}}} = 126x - 120x = 6x = 0,9 \to x = 0,15\)

\( \to m = 0,15.126 = 18,9{\text{ gam}}\)

\({n_{CaC{l_2}{\text{dư}}}} = 0,18 - 0,15 = 0,03{\text{ mol;}}{{\text{n}}_{NaCl}} = 2{n_{N{a_2}S{O_3}}} = 2x = 0,3{\text{ mol}}\)

\( \to {m_{CaC{l_2}{\text{ dư}}}} = 0,03.111 = 3,33{\text{gam;}}{{\text{m}}_{NaCl}} = 0,3.58,5 = 17,55{\text{ gam}}\)

BTKL: \({m_{dd\;{\text{Y}}}} = 111 + 0,9 = 111,9{\text{ gam}}\)

\( \to C{\% _{CaC{l_2}{\text{ du}}}} = \frac{{3,33}}{{111,9}} = 2,976\% ;C{\% _{NaCl}} = \frac{{17,55}}{{111,9}} = 15,68\% \)