Đáp án đúng: B
10,32 gam.
$\underbrace{{Al,}\,{C}{{{r}}_{{2}}}{{{O}}_{{3}}}}_{{19,52}\,{(g)}}\xrightarrow{{{{t}}^{{0}}}}\underbrace{{Al,}\,{A}{{{l}}_{{2}}}{{{O}}_{{3}}}{,}\,{Cr,}\,{C}{{{r}}_{{2}}}{{{O}}_{{3}}}}_{{19,52 (g) X}}\xrightarrow{{HCl}}\left\langle \begin{array}{l}{{{H}}_{{2}}}{:}\,{0,18}\,{mol}\\\underbrace{{A}{{{l}}^{{3+}}}{,}\,{C}{{{r}}^{{3+}}}{,}\,{C}{{{r}}^{{2+}}}{,}\,{C}{{{l}}^{{-}}}}_{{dd Y}}\xrightarrow{{NaOH}}\underbrace{{NaAl}{{{O}}_{{2}}}{,}\,{NaCr}{{{O}}_{{2}}}}_{{dd}}{+}\underbrace{{Cr(OH}{{{)}}_{{2}}}}_{{x}\,{(g)}\,\downarrow }\end{array} \right.$
– Khi cho hỗn hợp rắn X tác dụng với dung dịch chứa 0,96 mol HCl thì
$\xrightarrow{{BT:}\,{H}}{{{n}}_{{{{H}}_{{2}}}{O}}}{=}\frac{{{{n}}_{{HCl}}}{-2}{{{n}}_{{{{H}}_{{2}}}}}}{{2}}{=0,3}\,{mol}{{\Rightarrow }_{{C}{{{r}}_{{2}}}{{{O}}_{{3}}}}}{=}\frac{{{{n}}_{{{{H}}_{{2}}}{O}}}}{{3}}{=0,1}\,{mol}$
+ Xét hỗn hợp rắn ban đầu ta có:${{{n}}_{{Al}}}{=}\frac{{{{m}}_{{r}}}{-152}{{{n}}_{{C}{{{r}}_{{2}}}{{{O}}_{{3}}}}}}{{27}}{=0,16}\,{mol}$
– Xét quá trình nhiệt nhôm ta có:$\xrightarrow{{BT:}\,{e}}{{{n}}_{{C}{{{r}}_{{2}}}{{{O}}_{{3}}}{(pu)}}}{=}\frac{{3}{{{n}}_{{Al}}}{-2}{{{n}}_{{{{H}}_{{2}}}}}}{{2}}{=0,06}\,{mol}$
– Khi cho dung dịch Y tác dụng với NaOH dư thì:${{{m}}_{\downarrow }}{=86}{{{n}}_{{Cr(OH}{{{)}}_{{2}}}}}{=2}{.86}{.}{{{n}}_{{C}{{{r}}_{{2}}}{{{O}}_{{3}}}{(pu)}}}{=10,32}\,{gam}$
