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$M_xO_y\overset{+CO}{\rightarrow}xM+yCO_2$
$a ax ay$
$CO_2\overset{Ca(OH)_2}{\rightarrow}CaCO_3\downarrow +H_2O$
$ay 0,07 mol$
$\Rightarrow ay=0,07\Rightarrow m_M=4,06-0,07.16=2,94 gam\Rightarrow M=\frac{2,94}{ax } (1)$
$2M+2zHCl\rightarrow 2MCl_{z}+zH_2\uparrow$
$ax \frac{axz}{2}$
$\Rightarrow
\frac{axz}{2} =\frac{1,176}{22,4}=0,0525 mol\Rightarrow ax=\frac{0,105}{z} (2)$
$(1);(2)\Rightarrow M=28z$
Lập bảng $\Rightarrow n=2\Rightarrow M:Fe\Rightarrow ax:ay=x:y=3:4\Rightarrow Fe_3O_4$
$\Rightarrow n_{Fe_3O_4}=0,15 mol$
$
Fe_3O_4 +8HCl\rightarrow 2FeCl_3+FeCl_2$
$m=0,3.162,5+0,15.127=67,8 gam$
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Trả lời 03-01-13 06:01 PM
viet130480
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