Question

Answer 1

a) Gọi số mol CuO, MgO là a, b (mol)

=> 80a + 40b = 10 (1)

n_HCl = 0,1.3 = 0,3 (mol)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

               a---->2a------>a

            MgO + 2HCl --> MgCl₂ + H₂O

               b----->2b------>b

=> 2a + 2b = 0,3 (2)

(1)(2) => a = 0,1 (mol); b = 0,05 (mol)

=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1.80}{10}.100\%=80\%\\\%m_{MgO}=\dfrac{0,05.40}{10}.100\%=20\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}C_{M\left(CuCl_2\right)}=\dfrac{0,1}{0,1}=1M\\C_{M\left(MgCl_2\right)}=\dfrac{0,05}{0,1}=0,5M\end{matrix}\right.\)