Đáp án:
a) 0,5 M
b) 21,9g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
nC{O_2} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
nNaOH = {C_M} \times V = 1,5 \times 0,1 = 0,15\,mol\\
T = \dfrac{{nNaOH}}{{nC{O_2}}} = \dfrac{{0,15}}{{0,1}} = 1,5\\
1 < T < 2 \Rightarrow \text{ tạo ra 2 muối}\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
nN{a_2}C{O_3}(1) = nC{O_2}(1) = 0,075\,mol\\
nN{a_2}C{O_3}(2) = nC{O_2}(2) = 0,1 - 0,075 = 0,025\,mol\\
nN{a_2}C{O_3} = 0,075 - 0,025 = 0,05\,mol\\
nNaHC{O_3} = 0,025 \times 2 = 0,05\,mol\\
{C_M}NaHC{O_3} = {C_M}N{a_2}C{O_3} = \dfrac{{0,05}}{{0,1}} = 0,5M\\
b)\\
NaOH + HCl \to NaCl + {H_2}O\\
nHCl = nNaOH = 0,15\,mol\\
mHCl = n \times M = 0,15 \times 36,5 = 5,475g\\
m{\rm{dd}}HCl = \dfrac{{mct}}{{C\% }} = \dfrac{{5,475}}{{25\% }} = 21,9g
\end{array}\)
