Đáp án:
a)
\(\begin{array}{l}
{n_{N{a^ + }}} = 0,1mol\\
{n_{S{O_4}^{2 - }}} = 0,05mol
\end{array}\)
b)
\(\begin{array}{l}
{n_{B{a^{2 + }}}} = 0,1mol\\
{n_{C{l^ - }}} = 0,2mol
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
N{a_2}S{O_4} \to 2N{a^ + } + S{O_4}^{2 - }\\
{n_{N{a^ + }}} = 2 \times {n_{N{a_2}S{O_4}}} = 2 \times 0,05 = 0,1mol\\
{n_{S{O_4}^{2 - }}} = {n_{N{a_2}S{O_4}}} = 0,05mol\\
b)\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 5,2}}{{100}} = 20,8g\\
{n_{BaC{l_2}}} = \dfrac{{20,8}}{{208}} = 0,1mol\\
BaC{l_2} \to B{a^{2 + }} + 2C{l^ - }\\
{n_{B{a^{2 + }}}} = {n_{BaC{l_2}}} = 0,1mol\\
{n_{C{l^ - }}} = 2 \times {n_{BaC{l_2}}} = 2 \times 0,1 = 0,2mol
\end{array}\)
