Em tham khảo nha:

\(\begin{array}{l}
a)\\
{m_{Ca}} = 15,2 \times 26,316\%  = 4g\\
{m_{Fe}} = 15,2 - 4 = 11,2g\\
{n_{Ca}} = \dfrac{4}{{40}} = 0,1\,mol\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
2Ca + {O_2} \xrightarrow{t^0} 2CaO\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
{n_{{O_2}}} = 0,1 \times \dfrac{1}{2} + 0,2 \times \dfrac{2}{3} = \dfrac{{11}}{{60}}\,mol\\
{V_{{O_2}}} = \dfrac{{11}}{{60}} \times 22,4 = 4,11l\\
b)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Ca + 2HCl \to CaC{l_2} + {H_2}\\
{n_{HCl}} = 0,1 \times 2 + 0,2 \times 2 = 0,6\,mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
2)\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
\dfrac{{{n_{{H_2}}}}}{2} > {n_{{O_2}}} \Rightarrow {H_2} \text{ dư }\\
{n_{{H_2}O}} = 2{n_{{O_2}}} = 0,3\,mol\\
{m_{{H_2}O}} = 0,3 \times 18 = 5,4g
\end{array}\)

Giải thích các bước giải:

Câu 1:

$a)$

$m_{Ca}=15,2.26,316=4(g)$

$m_{Fe}=15,2-4=11,2(g)$

$n_{Ca}=\dfrac{4}{40}=0,1(mol)$

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$

$2Ca+O_2\xrightarrow{t^o}2CaO$ 

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$n_{O_2}=0,5n_{Ca}+\dfrac{2}{3}n_{Fe}=\dfrac{11}{60}(mol)$

$V_{O_2}=\dfrac{11}{60}.22,4=4,1(l)$

$b)$

$Ca+2HCl\to CaCl_2+H_2$

$Fe+2HCl\to FeCl_2+H_2$

$n_{HCl}=2(n_{Ca}+n_{Fe})=0,6(mol)$

$m_{HCl}=0,6.36,5=21,9(g)$

Câu 2:

$n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$

$n_{O_2}=\dfrac{3,36}{22,4}=0,15(mol)$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

$\dfrac{n_{H_2}}{2}>\dfrac{n_{O_2}}{1}\to H_2$ dư.

$n_{H_2O}=2n_{O_2}=0,3(mol)$

$m_{H_2O}=0,3.18=5,4(g)$