The solution of sodium meta aluminate on boiling with ammonium chloride gives a white precipitate of

Question

The solution of sodium meta aluminate on boiling with ammonium chloride gives a white precipitate of (A) Al 2 O 3 (B) AlCl 3 (C) Al ( OH )3 (D) ( NH 4)2 SO 4 ⋅ Al 2( SO 4)3

Tardigrade Admin · Accepted Answer

NH 4 Cl + NaAlO 2+ H 2 O arrow NaCl + Al ( OH )3+ NH 3

Q. The solution of sodium meta aluminate on boiling with ammonium chloride gives a white precipitate of

$A l_{2} O_{3}$

$A lC l_{3}$

$A l (O H)_{3}$

$(N H_{4})_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3}$

$N H_{4} Cl + N a A l O_{2} + H_{2} O \to N a Cl + A l (O H)_{3} + N H_{3}$

Q. The solution of sodium meta aluminate on boiling with ammonium chloride gives a white precipitate of

Al2​O3​

AlCl3​

Al(OH)3​

(NH4​)2​SO4​⋅Al2​(SO4​)3​

NH4​Cl+NaAlO2​+H2​O→NaCl+Al(OH)3​+NH3​

$A l_{2} O_{3}$

$A lC l_{3}$

$A l (O H)_{3}$

$(N H_{4})_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3}$

$N H_{4} Cl + N a A l O_{2} + H_{2} O \to N a Cl + A l (O H)_{3} + N H_{3}$