Đáp án:
Giải thích các bước giải:
Bài 2:
\(\begin{array}{l} a,\ PTHH:\\ CuO+CO\xrightarrow{t^o} Cu+CO_2↑\ (1)\\ PbO+CO\xrightarrow{t^o} Pb+CO_2↑\ (2)\\ CO_2+Ca(OH)_2\to CaCO_3↓+H_2O\ (3)\\ b,\ n_{CaCO_3}=\dfrac{11}{100}=0,11\ mol.\\ Theo\ pt\ (3):\ n_{CO_2}=n_{CaCO_3}=0,11\ mol.\\ Theo\ pt\ (1),(2):\ n_{CO}=n_{CO_2}=0,11\ mol.\\ ⇒V_{CO}=0,11\times 22,4=2,464\ lít.\\ c,\ \text{Gọi $n_{CuO}$ là a (mol), $n_{PbO}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\\ \left\{ \begin{array}{l}80a+223b=10,23\\a+b=0,11\end{array} \right.\ ⇒\left\{ \begin{array}{l}a=0,1\\b=0,01\end{array} \right.\\ ⇒\%m_{CuO}=\dfrac{0,1\times 80}{10,23}\times 100\%=78,2\%\\ \%m_{PbO}=\dfrac{0,01\times 223}{10,23}\times 100\%=21,8\%\end{array}\)
Bài 3:
\(\begin{array}{l} PTHH:\\ ZnO+2HCl\to ZnCl_2+H_2O\ (1)\\ FeO+2HCl\to FeCl_2+H_2O\ (2)\\ n_{HCl}=\dfrac{102,2\times 25\%}{36,5}=0,7\ mol.\\ \text{Gọi $n_{ZnO}$ là a (mol), $n_{FeO}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\\ \left\{ \begin{array}{l}81a+72b=26,55\\2a+2b=0,7\end{array} \right.\ ⇒\left\{ \begin{array}{l}a=0,15\\b=0,2\end{array} \right.\\ Theo\ pt\ (1):\ n_{ZnCl_2}=n_{ZnO}=0,15\ mol.\\ Theo\ pt\ (2):\ n_{FeCl_2}=n_{FeO}=0,2\ mol.\\ ⇒m_{ZnCl_2}=0,15\times 136=20,4\ g.\\ m_{FeCl_2}=0,2\times 127=25,4\ g.\end{array}\)
Bài 4:
\(\begin{array}{l} PTHH:\\ CuO+H_2SO_4\to CuSO_4+H_2O\ (1)\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\ (2)\\ \text{Ta có:}\ \dfrac{n_{CuCl_2}}{n_{FeCl_3}}=\dfrac{2}{3}\\ Theo\ pt\ (1):\ n_{CuO}=n_{CuCl_2}\\ Theo\ pt\ (2):\ n_{Fe_2O_3}=\dfrac{1}{2}n_{FeCl_3}\\ ⇒\dfrac{n_{CuO}}{2n_{Fe_2O_3}}=\dfrac{2}{3}\\ ⇒\dfrac{n_{CuO}}{n_{Fe_2O_3}}=\dfrac{2}{6}\\ ⇒\%n_{CuO}=\dfrac{2}{2+6}\times 100\%=25\%\\ \%n_{Fe_2O_3}=\dfrac{6}{2+6}\times 100\%=75\%\end{array}\)
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