632 g of sodium thiosulphate `(a_(2)S_(2)O_(3))` reacts with copper sulphate to form cuproc thiosulphate which is reduced by sodium thiosulphate to give cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothisosulphate `(Na_(4)[Cu_(6)(S_(2)O_(3))_(5)])`, (MW=1033)
`CuSO_(4) + Na_(2)S_(2)O_(3) rarr CuS_(2)O_(3) + Na_(2)SO_(4)" "` [very fast]
`2CuSO_(4)+Na_(2)S_(2)O_(3)+Na_(2)S_(2)O_(3) rarr Cu_(2)S_(2)O_(3) + Na_(2)S_(4)O_(6)`
`3Cu_(2)S_(2)O_(3) + 2Na_(2)S_(2)O_(3) rarr Na_(4)[Cu_(6)(S_(2)O_(3))_(5)]`
`" "` (Sodium cuprothisoulphate )
In this process , 0.2 mole of sodium cuprothiosulphate is formed .(O=16 , Na=23 , S=32)
Moles of sodium thiosulophate reacted and unreacted after the reaction are respectively,

Options

A. `3 & 2`
B. `2 & 3`
C. `2.2 & 1.8`
D. `1.8 & 2.2`

Explanation

To solve the problem, we need to calculate the moles of sodium thiosulphate (Na₂S₂O₃) that reacted and the moles that remained unreacted after the reaction with copper sulphate (CuSO₄) to form sodium cuprothiosulphate (Na₄[Cu₆(S₂O₃)₅]).

### Step-by-Step Solution:

1. **Calculate the moles of sodium thiosulphate (Na₂S₂O₃):**
- Given mass of sodium thiosulphate = 632 g
- Molecular weight (MW) of sodium thiosulphate (Na₂S₂O₃) = 158 g/mol
- Moles of Na₂S₂O₃ = Mass / Molecular Weight = 632 g / 158 g/mol = 4.0 moles

2. **Identify the moles of sodium cuprothiosulphate formed:**
- It is given that 0.2 moles of sodium cuprothiosulphate (Na₄[Cu₆(S₂O₃)₅]) are formed.

3. **Determine the stoichiometry of the reactions:**
- From the reactions provided:
- For the formation of sodium cuprothiosulphate, 3 moles of Cu₂S₂O₃ react with 2 moles of Na₂S₂O₃.
- Therefore, for every 0.2 moles of sodium cuprothiosulphate formed, the following amounts of sodium thiosulphate are consumed:
- From the reaction: \( 3 \text{ moles of } Cu_2S_2O_3 \) requires \( 2 \text{ moles of } Na_2S_2O_3 \)
- Thus, \( 0.2 \text{ moles of } Na_4[Cu_6(S_2O_3)_5] \) will require:
- \( 2 \times 0.2 = 0.4 \text{ moles of } Na_2S_2O_3 \) (from the last reaction)
- The previous reactions also contribute to the consumption of sodium thiosulphate:
- For \( 0.2 \text{ moles of } Cu_2S_2O_3 \), we have \( 3 \times 0.2 = 0.6 \text{ moles of } Na_2S_2O_3 \)
- For \( 0.2 \text{ moles of } CuS_2O_3 \), we have \( 2 \times 0.2 = 0.4 \text{ moles of } Na_2S_2O_3 \)

4. **Total moles of sodium thiosulphate reacted:**
- Total moles of Na₂S₂O₃ reacted = \( 0.4 + 0.6 + 0.4 = 1.4 \text{ moles} \)

5. **Calculate the unreacted moles of sodium thiosulphate:**
- Total moles of Na₂S₂O₃ initially = 4.0 moles
- Moles of Na₂S₂O₃ reacted = 1.4 moles
- Moles of Na₂S₂O₃ unreacted = Total moles - Reacted moles = \( 4.0 - 1.4 = 2.6 \text{ moles} \)

### Final Answer:
- Moles of sodium thiosulphate reacted: 1.4 moles
- Moles of sodium thiosulphate unreacted: 2.6 moles