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When $K_2Cr_2O_7$ crystals are heated with conc. HCl, the gas evolved is

(a) $O_2$
(b) $Cl_2$
(c) $CrO_2Cl_2$
(d) $HCl$

1 Answer

Answer: $Cl_2$
 
$K_2Cr_2O_7 + 8HCl \longrightarrow 2KCl + 2CrCl_3 + 4H_2O + 3[O]$
{$2HCl + [O] \longrightarrow H_2O + Cl_2$} $\times 3$
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$K_2Cr_2O_7 + 14HCl \longrightarrow 2KCl + 2CrCl_3 + 7H_2O + 3Cl_2 \uparrow$
answered Jan 28, 2014 by mosymeow_1
edited Jan 28, 2014
 

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