Đáp án:
\(\begin{array}{l}
a)\\
{m_{CaC{O_3}}} = 2,5g\\
b)\\
{V_{C{O_2}}} = 1,68l\\
c)\\
{m_{HCl}} = 1,825g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{CaC{O_3}}} = \dfrac{{10}}{{100}} = 0,1\,mol\\
{n_{HCl}} = \dfrac{{5,475}}{{36,5}} = 0,15\,mol\\
\dfrac{{0,1}}{1} > \dfrac{{0,15}}{2} \Rightarrow \text{ $CaCO_3$ dư}\\
{n_{CaC{O_3}}}\text{ dư} = 0,1 - \dfrac{{0,15}}{2} = 0,025\,mol\\
{m_{CaC{O_3}}}\text{ dư} = 0,025 \times 100 = 2,5g\\
b)\\
{n_{C{O_2}}} = \dfrac{{0,15}}{2} = 0,075\,mol\\
{V_{C{O_2}}} = 0,075 \times 22,4 = 1,68l\\
c)\\
{n_{HCl}} = 2{n_{CaC{O_3}}}\text{ dư} = 0,05\,mol\\
{m_{HCl}} = 0,05 \times 36,5 = 1,825g
\end{array}\)
