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geniusboy [140]
3 years ago
9

Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.

Chemistry
1 answer:
Contact [7]3 years ago
4 0

Answer:

0.42 M

Explanation:

The reaction that takes place is:

  • Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)

First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:

(200 mL = 0.200L)

  • 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄

Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:

  • 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂

Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:

  • 0.224 mol - 0.14 mol = 0.085 mol

Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:

  • 0.085 mol / 0.200 L = 0.42 M

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