6.100 mL of 0.100 M copper(II) nitrate is mixed in a beaker with 500 mL of 0.0100 M sodium hydroxide. How many moles of precipitate form? a. O millimoles b. 2.5 millimoles c. 5.0 millimoles d. 10 millimoles

Answers

Answer 1

As a result, the correct response is (b) 2.5 millimoles of precipitate form.

The balanced chemical equation for the reaction of sodium hydroxide and copper(II) nitrate is:

2NaOH + Cu(NO3)2 = Cu(OH)2 + 2NaNO3

One mole of copper(II) nitrate combines with two moles of sodium hydroxide to create one mole of copper(II) hydroxide, as shown by the equation.

We must first calculate the limiting reagent in the reaction before we can compute the amount of moles of precipitate that were produced.

Copper(II) nitrate concentration is indicated by:

C(V) = 0.100 mol/L (0.100 L) = 0.0100 mol for n(Cu(NO3)2).

You may find the sodium hydroxide concentration by:

C(V) = (0.0100 mol/L)(0.500 L) = 0.00500 mol for n(NaOH) and

The amount of sodium hydroxide is limited because it takes two moles of sodium hydroxide to react with one mole of copper(II) nitrate. This implies that the copper(II) nitrate will react with all of the sodium hydroxide present, and the amount of copper(II) hydroxide that results will depend on the sodium hydroxide present.

The formulas for: give the amount of copper(II) hydroxide that forms.

0.00500 mol/2 = 0.00250 mol for n(Cu(OH)2).

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Answer 2

As a result, the correct response is (b) 2.5 millimoles of precipitate form.

The balanced chemical equation for the reaction of sodium hydroxide and copper(II) nitrate is:2NaOH + Cu(NO3)2 = Cu(OH)2 + 2NaNO3One mole of copper(II) nitrate combines with two moles of sodium hydroxide to create one mole of copper(II) hydroxide, as shown by the equation.We must first calculate the limiting reagent in the reaction before we can compute the amount of moles of precipitate that were produced.Copper(II) nitrate concentration is indicated by:C(V) = 0.100 mol/L (0.100 L) = 0.0100 mol for n(Cu(NO3)2).You may find the sodium hydroxide concentration by:C(V) = (0.0100 mol/L)(0.500 L) = 0.00500 mol for n(NaOH) andThe amount of sodium hydroxide is limited because it takes two moles of sodium hydroxide to react with one mole of copper(II) nitrate. This implies that the copper(II) nitrate will react with all of the sodium hydroxide present, and the amount of copper(II) hydroxide that results will depend on the sodium hydroxide present.The formulas for: give the amount of copper(II) hydroxide that forms.0.00500 mol/2 = 0.00250 mol for n(Cu(OH)2).

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Answer:

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Answer:

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Consider this balanced chemical equation:
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