Bài 11:
1) \(n_{hh}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{CO_2}=a\left(mol\right)\\n_{SO_2}=b\left(mol\right)\end{matrix}\right.\)
=> a + b = 0,15 (1)
PTHH:
Ca(OH)2 + CO2 ---> CaCO3 + H2O
a---------->a
Ca(OH)2 + SO2 ---> CaSO3 + H2O
b--------->b
=> 100a + 120b = 17 (2)
Từ (1), (2) => a = 0,05; b = 0,1
=> \(\left\{{}\begin{matrix}\%V_{CO_2}=\dfrac{0,05}{0,15}.100\%=33,33\%\\\%V_{SO_2}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
2) \(n_{HCl}=0,1.3=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{ZnO}=b\left(mol\right)\end{matrix}\right.\)
=> 80a + 81b = 21 (1)
PTHH:
CuO + 2HCl ---> CuCl2 + H2O
a------>2a
ZnO + 2HCl ---> ZnCl2 + H2O
b------>2b
=> 2a + 2b = 0,3 (2)
Từ (1), (2) => a = -8,85; b = 9 (Đề sai à mà ra nghiệm âm?)
