Đáp án:
\(\begin{array}{l}
a)\\
{m_{MgS{O_4}}} = 12g\\
b)\\
C{\% _{BaC{l_2}}} = 10,4\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{H_2}S{O_4} + MgO \to MgS{O_4} + {H_2}O\\
MgS{O_4} + BaC{l_2} \to BaS{O_4} + MgC{l_2}\\
{n_{BaS{O_4}}} = \dfrac{m}{M} = \dfrac{{23,3}}{{233}} = 0,1mol\\
{n_{MgS{O_4}}} = {n_{BaS{O_4}}} = 0,1mol\\
{m_{MgS{O_4}}} = n \times M = 0,1 \times 120 = 12g\\
b)\\
{n_{BaC{l_2}}} = {n_{BaS{O_4}}} = 0,1mol\\
{m_{BaC{l_2}}} = n \times M = 0,1 \times 208 = 20,8g\\
C{\% _{BaC{l_2}}} = \dfrac{{20,8}}{{200}} \times 100\% = 10,4\%
\end{array}\)
