The number of mole of I₂ consumed by 0.65 mL of 0.15 M Na₂S₂O₃ is 4.88×10¯⁵ mole
The correct answer to the question is Option D. 4.88×10¯⁵ mole
We'll begin by calculating the number of mole of S₂O₃²¯ in 0.65 mL of 0.15 M Na₂S₂O₃. This can be obtained as follow:
Molarity of Na₂S₂O₃ = 0.15 M
Volume = 0.65 mL = 0.65 / 1000 = 6.5×10¯⁴ L
Mole of Na₂S₂O₃ =?
Mole = Molarity × Volume
Mole of Na₂S₂O₃ = 0.15 × 6.5×10¯⁴
Mole of Na₂S₂O₃ = 9.75×10¯⁵ mole
In solution,
Na₂S₂O₃(aq) —> 2Na⁺(aq) + S₂O₃²¯(aq)
From the balanced equation above,
1 mole of Na₂S₂O₃ contains 1 mole of S₂O₃²¯.
Therefore,
9.75×10¯⁵ mole of Na₂S₂O₃ will also contains 9.75×10¯⁵ mole of S₂O₃²¯.
Finally, we shall determine the number of mole of I₂ that will be consumed by the reaction. This is illustrated below:
2S₂O₃²¯ + I₂ —> 2I¯ + S₄O₆²¯
From the balanced equation above,
2 moles of S₂O₃²¯ consumes 1 mole of I₂.
Therefore,
9.75×10¯⁵ mole of S₂O₃²¯ will consume = [tex]\frac{9.75*10^{-5} }{2}\\\\[/tex] = 4.88×10¯⁵ mole of I₂.
Thus, the number of mole of I₂ consumed by 0.65 mL of 0.15 M Na₂S₂O₃ is 4.88×10¯⁵ mole
Option D. 4.88×10¯⁵ mole gives the correct answer to the question.
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