Question

How much of a 0.2M CuSO4, 0.4M NaI, and water is needed to prepare 12 mL of a 0.050 M I2 solution?

Answer 1

The balanced reaction for this is

2 CuSO4 + 4 NaI---> 2CuI + 2Na₂SO₄ + I₂

Moles of I₂ in 12 ml 0.050M I2

0.050× 0.012

= 0.0006 moles

Moles of CuSO4 needed = 0.0006 × 2

= 0.0012 moles

And 0.2 M solution volume will be

0.2 × V= 0.0012

= 0.006L

= 6ml

And moles of NaI required

= 0.0006 ×4

= 0.0024 moles

And volume of NaI required

= 0.4 ×V= 0.0024

= 0.006L

= 6 ml

So we have required moles and volume of 12 ml

(6+6)

So we will need

6 ml CuSO4 solution

6 ml NaI

And 0 ml water

I hope this helps if you have any query or want more detailed explanation feel free to ask in the comments section below.