Schweitzer's reagent is used in the rayon process. This reagent is prepared by passing NH 3 into solid blue vitriol. ( N H 4 ) 2 S O 4 NaOH N H 3 Bl u e v i t r i o l S c h w e i t

Question

Schweitzer's reagent is used in the rayon process. This reagent is prepared by passing $NH_{3}$ into solid blue vitriol.

$(N H_{4})_{2} S O_{4} NaOH N H_{3} Bl u e v i t r i o l S c h w e i t ze r^{'} sre a g e n t$

Determine the number of moles of Schweitzer's reagent formed when $528 g$ of pure $(N H_{4})_{2} S O_{4}$ .499 g of blue vitriol and excess of $NaOH$ are used in the above process (consider $100%$ yield in all cases)

(Atomic weight in $g / m o l^{- 1} : (H = 1, N = 14, O = 16, S = 32, N a = 23, C u = 63.5)$ ).

TestHub · Accepted Answer

Moles of ( NH 4 ​ ) 2 ​ SO 4 ​ = 4 Moles of CuSO 4 ​ ⋅ 5 H 2 ​ O = 2 ( NH 4 ​ ) 2 ​ SO 4 ​ + NaOH → 2 NH 3 ​ + Na 2 ​ SO 4 ​ + H 2 ​ O ∴ n NH 3 ​ ​ produced = 8 CuSO 4 ​ ⋅ 5 H 2 ​ O + 4 NH 3 ​ → [ Cu ( NH 3 ​ ) 4 ​ ] SO 4 ​ + 5 H 2 ​ O Mole: 2 8 0 0 moles of [ Cu ( NH 3 ​ ) 4 ​ ​ ] SO 4 ​ ​ produced = 2

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