Đáp án:
\(\begin{array}{l}
b)\\
\% {m_{CuO}} = 41,78\% \\
\% {m_{PbO}} = 58,22\% \\
c)\\
{m_C} = 0,9g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2CuO + C \to 2Cu + C{O_2}\\
2PbO + C \to 2Pb + C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
b)\\
{n_{CaC{O_3}}} = \dfrac{{7,5}}{{100}} = 0,075\,mol\\
\Rightarrow {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,075\,mol\\
hh:CuO(a\,mol),PbO(b\,mol)\\
80a + 223b = 19,15\\
0,5a + 0,5b = 0,075\\
\Rightarrow a = 0,1;b = 0,05\\
\% {m_{CuO}} = \dfrac{{0,1 \times 80}}{{19,15}} \times 100\% = 41,78\% \\
\% {m_{PbO}} = 100 - 41,78 = 58,22\% \\
c)\\
{n_C} = {n_{C{O_2}}} = 0,075\,mol\\
{m_C} = 0,075 \times 12 = 0,9g
\end{array}\)
