Chia 43,15 (g) hỗn hợp gồm Al, Fe2O3, Cr2O3 thành 2 phần.

* Hướng dẫn giải

Đáp án A

Phần (1): BT C \(\to {{n}_{C{{O}_{2}}}}+{{n}_{CO}}=0,06\,(mol)\to m=0,06.18.2=2,16(gam)\)

\(\begin{align} & BTKL\to {{m}_{Pi}}=2,16+25,41-0,06.28=25,89(gam)\to {{m}_{{{P}_{2}}}}=17,26(gam) \\ & \begin{matrix} A{{l}_{2}}{{O}_{3}}+2NaOH\to 2NaAl{{O}_{2}}+{{H}_{2}}O \\ Al+NaOH+{{H}_{2}}O\to NaAl{{O}_{2}}+\frac{3}{2}{{H}_{2}} \\ \end{matrix}\to {{n}_{Al(P2)}}={{n}_{NaOH}}=0,18(mol) \\ \end{align}\)

Phản ứng với HCl loãng nguội \(\to {{n}_{Fe}}={{n}_{{{H}_{2}}}}=0,06(mol)\)

\({{m}_{{{P}_{2}}}}=17,26(g)\to {{m}_{C{{r}_{2}}{{O}_{3}}}}=7,6(g)\to {{n}_{C{{r}_{2}}{{O}_{3}}}}=0,05(mol)\)

\(\begin{align} & \to {{n}_{O}}=0,05.3+0,06:2,3=0,24(mol) \\ & \to {{n}_{A{{l}_{2}}{{O}_{3}}}}=0,08(mol)\to {{n}_{Al}}=0,02(mol)\to {{V}_{{{H}_{2}}}}=1,5.0,02.22,4=0,672(l) \\ \end{align}\)