Đáp án:
\(\begin{array}{l}
6.\\
C{M_{NaOH}} = 1M\\
10.\\
{m_{{\rm{BaS}}{O_4}}} = 23,3g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6.\\
a)C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
b)\\
{n_{C{O_2}}} = 0,05mol\\
\to {n_{NaOH}} = 2{n_{C{O_2}}} = 0,1mol\\
\to C{M_{NaOH}} = \dfrac{{0,1}}{{0,1}} = 1M\\
10.\\
a)BaC{l_2} + {H_2}S{O_4} \to B{\rm{aS}}{O_4} + 2HCl\\
b)\\
{n_{{H_2}S{O_4}}} = 0,1mol\\
\to {n_{{\rm{BaS}}{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1mol\\
\to {m_{{\rm{BaS}}{O_4}}} = 23,3g
\end{array}\)
