Đáp án:

$\text{Giải thích các bước giải:}$

$\text{Bài 1:}$

$\text{a) 2Fe + $6H_{2}SO_{4}$→$Fe_{2}(SO_{4})_{3}$ + $3SO_{2}$ + $6H_{2}O$}$

$\text{b) Fe + $6HNO_{3}$→$Fe(NO_{3})_{3}$ + $3NO_{2}$ + $3H_{2}O$}$

$\text{c) Fe + $4HNO_{3}$→$Fe(NO_{3})_{3}$ + NO + $2H_{2}O$}$

$\text{d) 3FeS + $12HNO_{3}$→$Fe_{2}(SO_{4})_{3}$ + $Fe(NO_{3})_{3}$ + 9NO + $6H_{2}O$}$

Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\,2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
b)\,Fe + 6HN{O_3} \to Fe{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
c)\,Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
d)3\,FeS + 12HN{O_3} \to F{e_2}{(S{O_4})_3} + Fe{(N{O_3})_3} + 9NO + 6{H_2}O\\
2)\\
a)\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
b)\\
{n_{HCl}} = 0,1 \times 0,2 = 0,02\,mol\\
{n_{{H_2}O}} = \dfrac{{0,02}}{2} = 0,01\,mol\\
BTKL:\\
m = 2,3 + 0,02 \times 36,5 - 0,01 \times 18 = 2,85g\\
c)\\
{m_{{H_2}O}} = 0,01 \times 18 = 0,18g
\end{array}\)