Question

A reaction was ran and recovered 1.82 g of di(tetrabutylammonium)hexamolybdate. This gave a calculated yield of 91%. What is the minimum amount of sodium molybdate and TBAB required in grams?

Answer 1

. T B A B = Tetra butye Ammonium Bromide N(CH₂ CH₂ CH₂ Sodium molybdate= NG2 Modu TBAB + Sodium molybdate — di(tetra butyl ammonium) hexa molybdate + Sodium Bromide Required - 2 NlB4), Br chemical Equation. + Naz Molu - (M/B4) u'a Mo Oy + 2 Na Br 2 moles of NlBulu Br & 1 mole of Ma₂ Moly . required t form 1 mole of (N(Bu)ulz Moda to yield = Actual yield xloo Theoritical yield 91 - 1.82 x 100 n= 2g og Theoritical yield moles of (N(Bu) ud moda in ag = 644 g/mol. 4 29 mal = 3.1810-3 moles of (Ml Bula), This atleast this much of mole must produce. So, moles of N(Bulu Br used 2x moles of (N(Bu)u) ₂ Moly 2 * 3.1x10-3' moly

moles of TBAB (Ml Buty Br ) used = 6,2x10-3 moles molar mass of TBAB = 322 glmal Thus mass of TBAB required= 322 glmal x 6.2x10 mue (TBAB) M = 1,996 g udg Limilarly, moles of MQ, Molly used = 1 * mole og CMI I Buu), molly moles of & Na moou used = 1 x 3-1x 10-3 moles molae mass of Naamolu = 206 g/mol - This mass of ra,Moou required= 3.1x103 g mole x 206g1 mal Mq, mooy me = 0.638 )