Đáp án:
a) 3M
b) 15g
c) 42,98 ml
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
{n_{Ca{{(OH)}_2}}} = {n_{C{O_2}}} = 0,15\,mol\\
{C_M}Ca{(OH)_2} = \dfrac{{0,15}}{{0,05}} = 3M\\
b)\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15\,mol\\
{m_{CaC{O_3}}} = 0,15 \times 100 = 15g\\
c)\\
Ca{(OH)_2} + {H_2}S{O_4} \to CaS{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = {n_{Ca{{(OH)}_2}}} = 0,15\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,15 \times 98}}{{30\% }} = 49g\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{49}}{{1,14}} = 42,98ml
\end{array}\)