a)\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Tl 1 : 2 : 1 : 1 (mol)
BR 0,025 \(\leftarrow\) 0,05\(\rightarrow\)0,025 \(\rightarrow\)0,025(mol)
\(Đổi50ml=0,05l\)
\(\Rightarrow n_{HCl}=C_M.V_{dd}=0,05.1=0,05\left(mol\right)\)
b)m\(_{CuO}=n.M=0,025.80=2\left(g\right)\)
c)Taco V k thay đổi => V=0,05(l)
\(\Rightarrow C_M=\dfrac{n}{V}=\dfrac{0,025}{0,05}=0,5\left(M\right)\)
