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What is the mass in grams of precipitate formed when 250.0 mL of 0.360 M ZnCl₂ is mixed with excess KOH in the following chemical reaction?
ZnCl₂ (aq) + 2 KOH (aq) → Zn(OH)₂ (s) + 2 KCl (aq)

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Answer:

8.95 g

Explanation:

  • ZnCl₂ (aq) + 2 KOH (aq) → Zn(OH)₂ (s) + 2 KCl (aq)

First we calculate how many ZnCl₂ moles reacted, using the given concentration and volume:

  • 250 mL / 1000 = 0.250 L
  • 0.360 M * 0.250 L = 0.090 mol ZnCl₂

Then we convert ZnCl₂ moles into Zn(OH)₂ moles [ Zn(OH)₂ is the precipitate, as it is the solid]:

  • 0.090 mol ZnCl₂ * [tex]\frac{1molZn(OH)_2}{1molZnCl_2}[/tex] = 0.090 mol Zn(OH)₂

Finally we convert 0.090 Zn(OH)₂ moles into grams, using its molar mass:

  • 0.090 mol Zn(OH)₂ * 99.424 g/mol = 8.95 g

The mass of a substance is the product of the molar mass and the moles of the substance. The mass of precipitate formed in the reaction is 8.95 gm.

What is Mass?

The mass of the substance is the weight of the compound or the molecule present in the sample.

The balanced chemical reaction is shown as:

[tex]\rm ZnCl_{2} (aq) + 2 KOH (aq) \rightarrow Zn(OH)_{2} (s) + 2 KCl (aq)[/tex]

Given,

Volume (V) of zinc chloride = 0.250 L

Molarity of zinc chloride (M) = 0.360M

Moles of zinc chloride are calculated as:

[tex]\begin{aligned} \rm moles &= \rm molarity \times volume \\\\&= 0.360 \times 0.250\\\\&= 0.090 \;\rm mol\end{aligned}[/tex]

Moles of zinc hydroxide from zinc chloride are calculated as:

[tex]0.090\;\rm mol \; ZnCl_{2} \times \dfrac{ 1\;\rm mol \;\rm Zn(OH)_{2}}{1\;\rm mol\; ZnCl_{2}} = 0.090 \;\rm mol[/tex]

Mass is calculated as:

[tex]\begin{aligned} \rm mass &= \rm moles \times molar \; mass\\\\&= 0.090 \times 99.424 \\\\&= 8.95 \;\rm gm\end{aligned}[/tex]

Therefore, 8.95 gm precipitate is formed.

Learn more about mass here:

https://brainly.com/question/13834164

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