Đáp án:

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Giải thích các bước giải:

Bài 1:

\(\begin{array}{l}
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
a.\\
{n_{CuO}} = 0,025mol\\
 \to {n_{CuC{l_2}}} = {n_{CuO}} = 0,025mol\\
 \to {m_{CuC{l_2}}} = 3,375g
\end{array}\)

\(\begin{array}{l}
b.\\
{m_{{\rm{dd}}}} = {m_{CuO}} + {m_{{\rm{dd}}HCl}} = 102g\\
 \to C{\% _{CuC{l_2}}} = \dfrac{{3,375}}{{102}} \times 100\%  = 3,31\% 
\end{array}\)

Bài 2:

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
a)\\
{n_{{H_2}}} = 0,05mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,05mol\\
 \to {m_{Fe}} = 2,8g\\
 \to {m_{FeO}} = 10 - 2,8 = 7,2g\\
 \to {n_{FeO}} = 0,1mol\\
b)\\
{n_{HCl}} = 2{n_{Fe}} + 2{n_{FeO}} = 0,3mol\\
 \to C{M_{HCl}} = \dfrac{{0,3}}{{0,3}} = 1M
\end{array}\)

1.

` nCuO=0,025 (mol)`

`CuO + 2HCl -> CuCl2 + H_2O`

`a, nCuCl_2 = nCuO=0,025(mol)`

`  => mCuCl_2=3,375(g)`

`b, mdd = 2+100=102(g)`

`%CuCl_2 = (3,375)/(102).100=3,309%`

2.

` nH2=0,05 (mol)`

` FeO + 2HCl -> FeCl2 + H_2O`

` Fe + 2HCl -> FeCl2 + H_2`

`a, => nFe = nH2 =0,05(mol)`

`    => mFe = 2,8(g)`

`  => mFeO=7,2 (g)`

`    => nFeO=0,1 (mol)`

` b, nHCl = 2.nFeO + 2.nFe`

             `  = 2.0,1+2.0,05=0,3 (mol)`

` => CM_(HCl) = 0,3:0,3=1(M)`