Giải thích các bước giải:

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,6mol\\
 \to {n_{{H_2}S{O_4}}} > {n_{Fe}} \to {n_{{H_2}S{O_4}}}dư\\
 \to {n_{{H_2}S{O_4}(pt)}} = {n_{Fe}} = 0,2mol\\
 \to {n_{{H_2}S{O_4}(dư)}} = 0,4mol\\
 \to {n_{FeS{O_4}}} = {n_{Fe}} = 0,2mol\\
 \to C{M_{FeS{O_4}}} = \dfrac{{0,2}}{{0,2}} = 1M\\
 \to C{M_{{H_2}S{O_4}(du)}} = \dfrac{{0,4}}{{0,2}} = 2M\\
2.\\
{K_2}O + {H_2}O \to 2KOH\\
a)\\
{n_{{K_2}O}} = 0,1mol\\
 \to {n_{KOH}} = 2{n_{{K_2}O}} = 0,2mol\\
 \to C{M_{KOH}} = \dfrac{{0,2}}{{0,2}} = 1M\\
b)\\
{m_{KOH}} = 11,2g\\
{m_{{H_2}O}} = V \times D = 408,8g\\
 \to {m_{{\rm{dd}}}} = {m_{{K_2}O}} + {m_{{H_2}O}} = 418,2g\\
 \to C{\% _{KOH}} = \dfrac{{11,2}}{{418,2}} \times 100\%  = 2,68\% \\
3.\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{Mg}} = 0,075mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 50 \times 1,2 = 60g\\
a)\\
{n_{{H_2}S{O_4}}} = {n_{Mg}} = 0,075mol\\
 \to {m_{{H_2}S{O_4}}} = 7,35g\\
b)\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{7,35}}{{60}} \times 100\%  = 12,25\% \\
c)\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,075mol\\
 \to {m_{MgS{O_4}}} = 9g\\
 \to {m_{{\rm{dd}}}} = {m_{Mg}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 1,8 + 60 - 0,075 \times 2 = 61,65g\\
 \to C{\% _{MgS{O_4}}} = \dfrac{9}{{61,65}} \times 100\%  = 14,6\% 
\end{array}\)