Question

Answer 1

a. \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

a                        a       a

\(PbO+H_2\underrightarrow{t^o}Pb+H_2O\)

 b                   b         b

b. \(n_{H_2O}=\dfrac{0.9}{18}=0.05mol\)

Ta có: \(\left\{{}\begin{matrix}80a+223b=5.43\\a+b=0.05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.04\\b=0.01\end{matrix}\right.\)

\(\%m_{CuO}=\dfrac{0.04\times80\times100}{5.43}=58.93\%\)

\(\%m_{PbO}=100-58.93=41.07\%\)

c. m hỗn hợp rắn sau phản ứng \(=64a+207b=4.63g\)

\(\%m_{Cu}=\dfrac{0.04\times64\times100}{4.63}=55.3\%\)

\(\%m_{Pb}=100-55.3=44.7\%\)