Đáp án:

 Bạn tham khảo lời giải ở dưới nhé

Giải thích các bước giải:

\(\begin{array}{l}
6.\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
a)\\
{n_{C{O_2}}} = 0,15mol\\
 \to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15mol\\
 \to {m_{CaC{O_3}}} = 15g\\
 \to \% {m_{CaC{O_3}}} = \dfrac{{15}}{{20}} \times 100\%  = 75\% \\
 \to \% {m_{NaCl}} = 25\% \\
b)\\
{n_{HCl}} = 2{n_{CaC{O_3}}} = 0,3mol \to {m_{HCl}} = 10,95g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{10,95}}{{10\% }} \times 100\%  = 109,5g\\
7.\\
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
2NaOH + C{\rm{uS}}{O_4} \to N{a_2}S{O_4} + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
b)\\
{n_{N{a_2}O}} = 0,2mol\\
 \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,4mol\\
 \to {n_{Cu{{(OH)}_2}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol\\
 \to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
 \to {m_{CuO}} = 16g\\
c)\\
{n_{{\rm{CuS}}{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol \to {m_{{\rm{CuS}}{O_4}}} = 32g\\
 \to C{\% _{{\rm{CuS}}{O_4}}} = \dfrac{{32}}{{400}} \times 100\%  = 8\% \\
d)\\
CuO + CO \to Cu + C{O_2}\\
{n_{CO}} = {n_{CuO}} = 0,2mol\\
 \to {V_{CO}} = 4,48l
\end{array}\)