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To measure the amount of nickel in some industrial waste fluid, an analytical chemist adds 0.110 M sodium hydroxide (NaOH) solution to a 25.0 g sample of the fluid and collects the solid nickel(I) hydroxide (Ni (OH2) product. When no more Ni(OH)2 is produced, he filters, washes and weighs it, and finds that 343. mg has been produced The balanced chemical equation for the reaction is: Ni2+(aq) + 2NaOH(aq) Ni(OH)2(s) + 2 Na. (ag) R precipitation | | o acid-base o redox Og Dx10 Ar What kind of reaction is this? If you said this was a precipitation reaction, enter the chemical formula of the precipitate. If you said this was an acid-base reaction, enter the chemical formula of the reactant that is acting as the base. If you said this was a redox reaction, enter the chemical symbol of the element that is oxidized. Calculate the mass percent of Ni in the sample. Be sure your answer has the correct number of significant digits.

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Answer:

This is a precipitation reaction in which Ni(OH)₂ precipitates.

8.68%

Explanation:

Let's consider the following reaction.

Ni²⁺(aq) + 2 NaOH(aq) ⇄ Ni(OH)₂(s) + 2 Na⁺(aq)

This is a precipitation reaction in which Ni(OH)₂ precipitates.

We can establish the following relations:

  • The molar mass of Ni(OH)₂ is 92.71 g/mol.
  • 1 mole of Ni(OH)₂ is produced per 1 mole of Ni²⁺.
  • The molar mass of Ni²⁺ is 58.69 g/mol.

When 343 mg (0.343 g) of Ni(OH)₂ are collected, the mass of Ni²⁺ that reacted is:

[tex]0.343gNi(OH)_{2}.\frac{1molNi(OH)_{2}}{92.71gNi(OH)_{2}} .\frac{1molNi^{2+} }{1molNi(OH)_{2}} . \frac{58.69gNi^{2+}}{1molNi^{2+}} =0.217gNi^{2+}[/tex]

The mass percent of nickel in the 25.0g-sample is:

[tex]\frac{0.217g}{2.50g}.100\%=8.68\%[/tex]

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