Em tham khảo nha :

\(\begin{array}{l}
7)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
{V_{{H_2}}} = 22,4 \times n = 22,4 \times 0,2 = 4,48l\\
8)\\
CaC{O_3} \to CaO + C{O_2}\\
{n_{CaC{O_3}}} = \dfrac{m}{M} = \dfrac{{10}}{{100}} = 0,1mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,1mol\\
{V_{C{O_2}}} = 22,4 \times n = 22,4 \times 0,1 = 2,24l\\
9)\\
a)\\
4Fe + 3{O_2} \to 2F{e_2}{O_3}\\
{n_{Fe}} = \dfrac{m}{M} = \frac{{16,8}}{{56}} = 0,3mol\\
 \Rightarrow {n_{{O_2}}} = \dfrac{3}{4}{n_{Fe}} = 0,225mol\\
{V_{{O_2}}} = 22,4 \times n = 22,4 \times 0,225 = 5,04l\\
b)\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe}} = 0,15mol\\
{n_{{H_2}S{O_4}}} = 3{n_{F{e_2}{O_3}}} = 0,45mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,45 \times 98 = 44,1g
\end{array}\)

Đáp án:

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