Đáp án đúng: D
5,416l.
Trong 15,8 gam A có ${{m}_{KMn{{O}_{4}}}}\,=\,\,13,904\,\,gam$ và${{m}_{Mn{{O}_{2}}}}\,=\,1,896\,gam$
$\Rightarrow {{n}_{KMn{{O}_{4}}}}\,=\,0,088\,mol;\,{{n}_{Mn{{O}_{2}}}}\,=\,0,0218\,mol$
${{m}_{dd\,HCl}}=\,166,6\,gam\,\Rightarrow \,{{m}_{HCl}}\,=\,63,6412\,gam\,\Rightarrow {{n}_{HCl}}\,=\,1,7346\,mol$
$2KMn{{O}_{4}}$
+
$16HCl$
→
$2KCl$
+
$2MnC{{l}_{2}}$
+
$5C{{l}_{2}}$
+
$8{{H}_{2}}O$
0,088
→
0,704
→
0,22
$Mn{{O}_{2}}$
+
$4HCl$
→
$MnC{{l}_{2}}$
+
$C{{l}_{2}}$
+
$2{{H}_{2}}O$
0,0218
→
0,0872
→
0,0218
⇒ HCl dư
$\Rightarrow {{V}_{C{{l}_{2}}}}\,=\,5,416\,lit$
