consider the reaction that occurs when 7.5 ml of 1.2 m barium nitrite is mixed with 10.0 ml of 0.60 m sodium sulfate. a) How many grams of barium sulfate are produced if this reaction runs with a 100% yield? b) what ions remain in solution? c)what concentration of ions remain in the solution?

Answers

Answer 1

a) 1.40 g of [tex]BaSO_{4}[/tex] are produced if the reaction runs with a 100% yield.

b) Na+ and [tex]NO_{3-}[/tex] are the ions remain in solution

c) the concentration of remaining Na+ ions is 0.012 M, and [tex]NO_{3-}[/tex]ions is 0.018 M.

a) The balanced equation for the reaction is:

[tex]Ba(NO_{3}){2}[/tex](aq) + [tex]Na_{2}SO_{4}[/tex] (aq) → [tex]BaSO_{4}[/tex] (s) + [tex]2NaNO_{3}[/tex] (aq)

From the equation, we can see that one mole of barium nitrite reacts with one mole of sodium sulfate to produce one mole of barium sulfate. Therefore, we need to calculate the number of moles of barium nitrite and sodium sulfate to determine the limiting reagent and the theoretical yield.

Number of moles of [tex]BaNO_{3}{2}[/tex] = 1.2 M x (7.5/1000) L = 0.009 moles

Number of moles of [tex]Na_{2}SO_{4}[/tex] = 0.60 M x (10.0/1000) L = 0.006 moles

Since [tex]Na_{2}SO_{4}[/tex] is the limiting reagent, it will be completely consumed in the reaction. The theoretical yield of [tex]BaSO_{4}[/tex]can be calculated as:

Theoretical yield of [tex]BaSO_{4}[/tex] = 0.006 moles x 233.4 g/mol (molar mass of [tex]BaSO_{4}[/tex]) = 1.40 g

Therefore, 1.40 g of [tex]BaSO_{4}[/tex]are produced if the reaction runs with a 100% yield.

b) The ions that remain in solution after the reaction are Na+ and [tex]NO_{3-}[/tex].

c) To calculate the concentration of remaining ions, we need to determine how much of each ion is present in solution before the reaction. From the balanced equation, we can see that one mole of [tex]Na_{2}SO_{4}[/tex]produces two moles of Na+ and one mole of [tex]SO_{42-}[/tex]. Therefore, the initial concentration of Na+ is:

Initial concentration of Na+ = 0.60 M x (10.0/1000) L x 2 = 0.012 M

Similarly, the initial concentration of [tex]NO_{3-}[/tex] is:

Initial concentration of [tex]NO_{3-}[/tex] = 1.2 M x (7.5/1000) L x 2 = 0.018 M

After the reaction, all of the Na+ ions remain in solution, while all of the [tex]NO_{3-}[/tex] ions form [tex]NaNO_{3}[/tex] and remain in solution. Therefore, the concentration of remaining Na+ ions is 0.012 M, and the concentration of remaining [tex]NO_{3-}[/tex] ions is 0.018 M.

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Answer:

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Answer:

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Answer:

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Answer:

Explanation:

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Answers

Answer:

5.3g of MgO would be produced

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