Em tham khảo nha :
Em xem lại đề câu 2 nha
\(\begin{array}{l}
1)\\
a)\\
F{e_x}{O_y} + 2yHCl \to xFe + y{H_2}O\\
b)\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{5,475}}{{36,5}} = 0,15mol\\
{n_{F{e_x}{O_y}}} = \dfrac{{{n_{HCl}}}}{{2y}} = \dfrac{{0,075}}{y}mol\\
{M_{F{e_x}{O_y}}} = \dfrac{m}{n} = 4:\dfrac{{0,075}}{y} = \dfrac{{160}}{3}y\\
\Rightarrow 56x + 16y = \dfrac{{160}}{3}y\\
y = 3 \Rightarrow x = \dfrac{{160 - 16 \times 3}}{{56}} = 2\\
\Rightarrow CTHH:F{e_2}{O_3}\\
2)\\
a)\\
\text{Hợp chất X gồm 3 nguyên tố : Cacbon , Hidro , Oxi}\\
b)\\
CTHH:{C_x}{H_y}\\
{M_X} = {M_{{O_2}}} = 32dvC\\
{n_{C{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{{H_2}O}} = \dfrac{{7,2}}{{18}} = 0,4mol\\
{n_C} = {n_{C{O_2}}} = 0,2mol\\
{n_H} = 2{n_{{H_2}O}} = 0,8mol\\
{m_O} = 6,4 - 0,2 \times 12 - 0,8 \times 1 = 3,2g\\
{n_O} = \dfrac{{3,2}}{{16}} = 0,2mol\\
{n_X} = \dfrac{{6,4}}{{32}} = 0,2mol\\
x = \dfrac{{{n_C}}}{{{n_X}}} = \dfrac{{0,2}}{{0,1}} = 1\\
y = \dfrac{{{n_H}}}{{{n_X}}} = \dfrac{{0,8}}{{0,2}} = 4\\
z = \dfrac{{{n_O}}}{{{n_X}}} = \dfrac{{0,2}}{{0,1}} = 1\\
\Rightarrow CTPT:C{H_4}O
\end{array}\)
