4)

Ta có:

\({n_{C{O_2}}} = \frac{{0,672}}{{22,4}} = 0,03{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,2.0,2 = 0,04{\text{ mol}}\)

\( \to \frac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \frac{{0,04}}{{0,03}} = 1,333\)

Vậy sản phẩm tạo ra gồm 2 muối.

\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)

\(NaOH + C{O_2}\xrightarrow{{}}NaHC{O_3}\)

\( \to {n_{N{a_2}C{O_3}}} = {n_{NaOH}} - {n_{C{O_2}}} = 0,04 - 0,03 = 0,01{\text{ mol}}\)

\( \to {n_{NaHC{O_3}}} = 0,03 - 0,01 = 0,02{\text{ mol}}\)

\( \to {m_{muối}} = {m_{NaHC{O_3}}} + {m_{N{a_2}C{O_3}}} = 0,02.84 + 0,01.106 = 2,74{\text{ gam}}\)

5)

Phản ứng xảy ra:

\(CaC{{\text{O}}_3}\xrightarrow{{{t^o}}}CaO + C{O_2}\)

\(BaC{{\text{O}}_3}\xrightarrow{{{t^o}}}BaO + C{O_2}\)

\(Ba{(OH)_2} + C{O_2}\xrightarrow{{{t^o}}}BaC{{\text{O}}_3} + {H_2}O\)

Ta có:

\({n_{BaC{{\text{O}}_3}}} = \frac{{15,76}}{{197}} = 0,08{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)

Bảo toàn khối lượng:

\( \to a = {m_X} + {m_{C{O_2}}} = 3,68 + 0,08.44 = 7,2{\text{ gam}}\)