Đáp án:
\(\begin{array}{l}
a)\\
m = 50,1g\\
b)\\
{m_{{H_2}S{O_4}}} = 29,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Ba + {H_2}S{O_4} \to BaS{O_4} + {H_2}(1)\\
BaC{O_3} + {H_2}S{O_4} \to BaS{O_4} + C{O_2} + {H_2}O(2)\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{BaS{O_4}}} = \dfrac{{69,9}}{{233}} = 0,3\,mol\\
{n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,15\,mol\\
{n_{BaS{O_4}(2)}} = {n_{C{O_2}}} = 0,15\,mol\\
\Rightarrow {n_{BaS{O_4}(1)}} = 0,3 - 0,15 = 0,15\,mol\\
{n_{Ba}} = {n_{BaS{O_4}(1)}} = 0,15\,mol\\
m = {m_{Ba}} + {m_{BaC{O_3}}} = 0,15 \times 137 + 0,15 \times 197 = 50,1g\\
b)\\
{n_{{H_2}S{O_4}}} = 0,15 + 0,15 = 0,3\,mol\\
{m_{{H_2}S{O_4}}} = 0,3 \times 98 = 29,4g
\end{array}\)
