Question

E solution A O 1 1 1 19 15g of acetyl acetone (CH3O₂). 20nh H₂O 4 mL of 5m ammonia add B to A 1.75g of Al2(SO4)3. 14420) 15mL ice cold H₂O 7 solution B (Murral phi) or 1.09 Al (CGH10₂), Yeild moleculor = 324 g moll weight

write balance a equation for this reaction and calculate the %yield of the product.

Answer 1

1_ ta,

balanced chemical equation

6NH₄⁺+6C₅H₇O₂^-+Al₂(SO₄)₃=2Al(C₅H₇O₂)₃+3(NH₄)₂SO4

Molecular weight of C₅H₇O₂^-=99 gm/mol

Molecular weight of Al(C₅H₇O₂)₃=324 gm/mol.

(6$\times$99) gm C₅H₇O₂^- should yield (2$\times$324) gm of Al(C₅H₇O₂)₃

_{1.5 gm 5}H₇O₂^- should yield $\frac{2\times 324 \times 1.5}{6 \times 99}$ gm of Al(C₅H₇O₂)₃

                                     = 1.636 gm Al(C₅H₇O₂)₃

_{Theoretical yield =1.636} gm

Actual yield =1.09 gm

percentage of yield =$\frac{1.09\times 100}{1.636}$=66.626%