PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a______2a______a (mol)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b_____2b______b (mol)
Ta lập được HPT: \(\left\{{}\begin{matrix}40a+80b=20\\2a+2b=0,2\cdot3=0,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,1\cdot40}{20}\cdot100\%=20\%\\\%m_{CuO}=80\%\\m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{CuCl_2}=0,2\cdot135=27\left(g\right)\end{matrix}\right.\)
a)
$MgO + 2HCl \to MgCl_2 + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2O$
b)
Gọi n MgO = a(mol) ; n CuO =b(mol)
=> 40a + 80b = 20(1)
Theo PTHH :
n HCl = 2a + 2b = 0,2.3 =0,6(2)
Từ (1)(2) suy ra a = 0,1 ; b = 0,2
%m MgO = 0,1.40/20 .100% = 20%
%m CuO = 100% -20% = 80%
c)
n MgCl2 = a = 0,1(mol)
n CuCl2 = b = 0,2(mol)
=> m = 0,1.95 + 0,2.135 = 36,5 gam
