Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

Bài 1:

\(\begin{array}{l}
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{98 \times 40}}{{100 \times 98}} = 0,4mol\\
 \to {n_{CuO}} = {n_{{H_2}S{O_4}}} = 0,4mol\\
 \to m = {m_{CuO}} = 32g\\
{n_{CuS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,4mol\\
 \to {m_{CuS{O_4}}} = 64g\\
{m_{dd}} = {m_{CuO}} + {m_{{H_2}S{O_4}}}dd = 32 + 98 = 130g\\
 \to C{\% _{CuS{O_4}}} = \dfrac{{64}}{{130}} \times 100\%  = 49,23\% 
\end{array}\)

Bài 2:

\(\begin{array}{l}
F{{\rm{e}}_2}{O_3} + 6HCl \to 2F{\rm{e}}C{l_3} + 3{H_2}O\\
{n_{F{{\rm{e}}_2}{O_3}}} = 0,1mol\\
{n_{HCl}} = \dfrac{{146 \times 20}}{{100 \times 36,5}} = 0,8mol\\
 \to {n_{F{{\rm{e}}_2}{O_3}}} < \dfrac{{{n_{HCl}}}}{6}
\end{array}\)

Suy ra dung dịch A có HCl dư 

\(\begin{array}{l}
 \to {n_{F{\rm{e}}C{l_3}}} = 2{n_{F{{\rm{e}}_2}{O_3}}} = 0,2mol\\
 \to {n_{HCl}}dư= 0,8 - 6{n_{F{{\rm{e}}_2}{O_3}}} = 0,2mol\\
 \to {m_A} = {m_{F{\rm{e}}C{l_3}}} + {m_{HCl}}dư= 39,8g\\
{m_{{\rm{dd}}}} = {m_{F{{\rm{e}}_2}{O_3}}} + {m_{HCl}}{\rm{dd = 16 + 146 = 162g}}\\
C{\% _{F{\rm{e}}C{l_3}}} = \dfrac{{32,5}}{{162}} \times 100\%  = 20,06\% \\
C{\% _{HCl}} = \dfrac{{7,3}}{{162}} \times 100\%  = 4,51\% 
\end{array}\)

Bạn tham khảo:

$1/$

$a/$

$CuO+H_2SO_4 \to CuSO_4+H_2O$

$n_{H_2SO_4}=\frac{98.40\%}{98}=0,4(mol)$

$n_{CuO}=n_{H_2SO_4}=0,4(mol)$
$m_{CuO}=0,4.80=32(g)$
$b/$

$n_{CuSO_4}=n_{H_2SO_4}=0,4(mol)$

$C\%_{CuSO_4}=\frac{0,4.160}{32+98}.100=49,23\%$

$2/$

$Fe_2O_3+6HCl \to 2FeCl_3+3H_2O$

$a/$

$n_{Fe_2O_3}=0,1(mol)$

$n_{HCl}=\frac{146.20\%}{36,5}=0,8(mol)$

$0,1 < \frac{0,8}{6}=0,133$

$Fe_2O_3$ hết; $HCl$ dư

$n_{HCl}=0,1.6=0,6(mol)$

$m_{HCl}=(0,8-0,6).36,5=7,3(g)$

$n_{FeCl_3}=0,1.2=0,2(mol)$
$m_{FeCl_3}=0,2.162,5=32,5(g)$

$b/$

$C\%_{HCl}=\frac{7,3}{16+146}100\%=4,5\%$

$C\%_{FeCl_3}=\frac{32,5}{16+146}.100\%=20,6\%$