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labwork [276]
2 years ago
8

Write balanced net ionic equations for the three reactions carried out in Part A given that the sulfur-containing product is the

S2O3
Chemistry
1 answer:
Alik [6]2 years ago
3 0

Answer:

Please find the complete question in the attached file.

Explanation:

Balanced equation: 8KMnO_4+14NaOH +Na_2S_2O_3 \longrightarrow 8Na_2MnO_4 +2K_2SO_4 +5H_2O +4KOH

Ionic equation:

8K^{+} + 8MnO_4^{-}+14Na^{+}+ 14OH^- +2Na^+ S_2O_3^{-}\longrightarrow16Na^+8MnO_4^- +4K^+ + 2SO_4 +5H_2O+4K^+ +4OH^-

Net Ionic equation:

10 OH^- +S_2O_3^- \longrightarrow 2SO_4^- +5H_2O

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Answer:

62.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

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From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:

From the balanced equation above,

2 moles of C₈H₁₈ reacted with 25 moles of O₂.

Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.

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