Đáp án:
\(\begin{array}{l}
b)\\
{V_{{\rm{dd}}Ca{{(OH)}_2}}} = 0,03l\\
c)\\
\% {V_{CO}} = 40\% \\
\% {V_{C{O_2}}} = 60\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
CuO + CO \to Cu + C{O_2}\\
b)\\
{n_{CaC{O_3}}} = \dfrac{{1,5}}{{100}} = 0,015\,mol\\
{n_{Ca{{(OH)}_2}}} = {n_{CaC{O_3}}} = 0,015\,mol\\
{V_{{\rm{dd}}Ca{{(OH)}_2}}} = \dfrac{{0,015}}{{0,5}} = 0,03l\\
c)\\
{n_{CO}} = {n_{Cu}} = \dfrac{{0,64}}{{64}} = 0,01\,mol\\
{{\rm{n}}_{C{O_2}}} = {n_{CaC{O_3}}} = 0,015\,mol\\
\% {V_{CO}} = \dfrac{{0,01}}{{0,015 + 0,01}} \times 100 = 40\% \\
\% {V_{C{O_2}}} = 100 - 40 = 60\%
\end{array}\)