Chemical Forums

Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Win,odd Dhamnekar on April 08, 2023, 03:13:34 AM

Title: Occurrences, preparations and properties of Noble gases
Post by: Win,odd Dhamnekar on April 08, 2023, 03:13:34 AM

Question:

Basic solutions of Na₄XeO₆ are powerful oxidants. What mass of Mn(NO₃)₂•6H₂O reacts with 125.0 mL of a 0.1717 M basic solution of Na₄XeO₆ that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate?

My Answer:

To solve this problem, we will follow these steps:

1. Write the balanced chemical equation for the reaction.
2. Find the moles of [itex]Na_4XeO_6[/itex] in the given solution.
3. Determine the stoichiometric ratio between [itex]Mn(NO_3)_2•6H_2O[/itex] and [itex]Na_4XeO_6[/itex].
4. Calculate the mass of [itex]Mn(NO_3)_2•6H_2O[/itex].

Step 1: Write the balanced chemical equation for the reaction.

The balanced chemical equation is:

$$2(MnNO_3)_2•6H_2O + 3Na_4XeO_6  → 3Xe + 4NaOH + 4NaMnO_4 + 4NaNO_3 + 10H_2O$$

Step 2: Find the moles of [itex]Na_4XeO_6[/itex] in the given solution.

Moles of [itex]Na_4XeO_6[/itex] = Molarity × Volume

Moles of [itex]Na_4XeO_6[/itex] = 0.1717 mol/L × 0.125 L = 0.0214625 mol

Step 3: Determine the stoichiometric ratio between [itex](MnNO_3)_2•6H_2O[/itex] and [itex]Na_4XeO_6[/itex].

From the balanced chemical equation, we have:

2 moles of [itex]Mn(NO_3)_2•6H_2O[/itex] react with 3 moles of [itex]Na_4XeO_6[/itex]

So, moles of [itex](MnNO_3)_2•6H_2O[/itex] = (2/3) × moles of [itex]Na_4XeO_6[/itex]

Moles of [itex]Mn(NO_3)_2•6H_2O[/itex] = (2/3) × 0.0214625 mol = 0.0014308 mol

Step 4: Calculate the mass of [itex](MnNO_3)_2•6H_2O[/itex].

First, we need to find the molar mass of [itex](MnNO_3)_2•6H_2O[/itex]:

Molar mass of [itex](MnNO_3)_2•6H_2O[/itex] = 341.974 g/mol

Now, we can calculate the mass of [itex](MnNO_3)_2•6H_2O[/itex]:

Mass of [itex](MnNO_3)_2•6H_2O[/itex] = moles × molar mass
= 0.0014308 mol × 341.974 g/mol = 4.89308 g

So, 4.89308 g of [itex](MnNO_3)_2•6H_2O[/itex] reacts with 125.0 mL of the 0.1717 M basic solution of [itex]Na_4XeO_6[/itex].

Is my answer given above correct?

Title: Re: Occurrences, preparations and properties of Noble gases
Post by: Hunter2 on April 08, 2023, 07:01:24 AM

It start the equation is wrong, not balanced

Error with the brackets at Manganese nitrate.

Title: Re: Occurrences, preparations and properties of Noble gases
Post by: Win,odd Dhamnekar on April 08, 2023, 09:36:03 AM

Quote from: Hunter2 on April 08, 2023, 07:01:24 AM

It start the equation is wrong, not balanced

Error with the brackets at Manganese nitrate.

Then what is the correct balanced chemical equation?

Title: Re: Occurrences, preparations and properties of Noble gases
Post by: Hunter2 on April 08, 2023, 09:46:14 AM

I am talking about (MnNO3)2*6 H2O, but it is Mn(NO3)2*6 H2O
So there is only one Mn involved.

Oxidation is Mn2+ + 8 OH-  => MnO4- + 4 H2O + 5 e-
Reduction XeO6 4- + 6 H2O + 8 e- => Xe + 12 OH-

The Rest calculate your self.

Title: Re: Occurrences, preparations and properties of Noble gases
Post by: billnotgatez on April 08, 2023, 10:29:59 AM

Quote from: Hunter2 on April 08, 2023, 09:46:14 AM

I am talking about (MnNO3)2*6 H2O, but it is Mn(NO3)2*6 H2O
.

It took me several tries to catch it
(MnNO3)2*6 H2O
Mn(NO3)2*6 H2O

good catch

Title: Re: Occurrences, preparations and properties of Noble gases
Post by: Win,odd Dhamnekar on April 09, 2023, 02:36:45 AM

Quote from: Win,odd Dhamnekar on April 08, 2023, 03:13:34 AM

Question:

Basic solutions of Na₄XeO₆ are powerful oxidants. What mass of Mn(NO₃)₂•6H₂O reacts with 125.0 mL of a 0.1717 M basic solution of Na₄XeO₆ that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate?

My Answer:

To solve this problem, we will follow these steps:

1. Write the balanced chemical equation for the reaction.
2. Find the moles of [itex]Na_4XeO_6[/itex] in the given solution.
3. Determine the stoichiometric ratio between [itex]Mn(NO_3)_2•6H_2O[/itex] and [itex]Na_4XeO_6[/itex].
4. Calculate the mass of [itex]Mn(NO_3)_2•6H_2O[/itex].

Step 1: Write the balanced chemical equation for the reaction.

The balanced chemical equation is:

$$2(MnNO_3)_2•6H_2O + 3Na_4XeO_6  → 3Xe + 4NaOH + 4NaMnO_4 + 4NaNO_3 + 10H_2O$$

Step 2: Find the moles of [itex]Na_4XeO_6[/itex] in the given solution.

Moles of [itex]Na_4XeO_6[/itex] = Molarity × Volume

Moles of [itex]Na_4XeO_6[/itex] = 0.1717 mol/L × 0.125 L = 0.0214625 mol

Step 3: Determine the stoichiometric ratio between [itex](MnNO_3)_2•6H_2O[/itex] and [itex]Na_4XeO_6[/itex].

From the balanced chemical equation, we have:

2 moles of [itex]Mn(NO_3)_2•6H_2O[/itex] react with 3 moles of [itex]Na_4XeO_6[/itex]

So, moles of [itex](MnNO_3)_2•6H_2O[/itex] = (2/3) × moles of [itex]Na_4XeO_6[/itex]

Moles of [itex]Mn(NO_3)_2•6H_2O[/itex] = (2/3) × 0.0214625 mol = 0.0014308 mol

Step 4: Calculate the mass of [itex](MnNO_3)_2•6H_2O[/itex].

First, we need to find the molar mass of [itex](MnNO_3)_2•6H_2O[/itex]:

Molar mass of [itex](MnNO_3)_2•6H_2O[/itex] = 341.974 g/mol

Now, we can calculate the mass of [itex](MnNO_3)_2•6H_2O[/itex]:

Mass of [itex](MnNO_3)_2•6H_2O[/itex] = moles × molar mass
= 0.0014308 mol × 341.974 g/mol = 4.89308 g

So, 4.89308 g of [itex](MnNO_3)_2•6H_2O[/itex] reacts with 125.0 mL of the 0.1717 M basic solution of [itex]Na_4XeO_6[/itex].

Is my answer given above correct?

My corrected Answer:

To solve this problem, we will follow these steps:

1. Write the balanced chemical equation for the reaction.
2. Find the moles of [itex]Na_4XeO_6[/itex] in the given solution.
3. Determine the stoichiometric ratio between [itex]Mn(NO_3)_2•6H_2O[/itex] and [itex]Na_4XeO_6[/itex].
4. Calculate the mass of [itex]Mn(NO_3)_2•6H_2O[/itex].

Step 1: Write the balanced chemical equation for the reaction.

The balanced chemical equation is:

$$8Mn(NO_3)_2•6H_2O + 5Na_4XeO_6 + 4NaOH → 5Xe  + 8NaMnO_4 + 16NaNO_3 + 50H_2O$$

Step 2: Find the moles of [itex]Na_4XeO_6[/itex] in the given solution.

Moles of [itex]Na_4XeO_6[/itex] = Molarity × Volume

Moles of [itex]Na_4XeO_6[/itex] = 0.1717 mol/L × 0.125 L = 0.0214625 mol

Step 3: Determine the stoichiometric ratio between [itex]Mn(NO_3)_2•6H_2O[/itex] and [itex]Na_4XeO_6[/itex].

From the balanced chemical equation, we have:

8 moles of [itex]Mn(NO_3)_2•6H_2O[/itex] react with 5 moles of [itex]Na_4XeO_6[/itex]

So, moles of [itex]Mn(NO_3)_2•6H_2O[/itex] = (8/5) × moles of [itex]Na_4XeO_6[/itex]

Moles of [itex]Mn(NO_3)_2•6H_2O[/itex] = (8/5) × 0.0214625 mol = 0.03434 mol

Step 4: Calculate the mass of [itex]Mn(NO_3)_2•6H_2O[/itex].

First, we need to find the molar mass of [itex]Mn(NO_3)_2•6H_2O[/itex]:

Molar mass of [itex]Mn(NO_3)_2•6H_2O[/itex] = 287.036 g/mol

Now, we can calculate the mass of [itex]Mn(NO_3)_2•6H_2O[/itex]:

Mass of [itex]Mn(NO_3)_2•6H_2O[/itex] = moles × molar mass
= 0.03434 g mol × 287.036 g/mol = 9.857 g

So, 9.857 g of [itex](MnNO_3)_2•6H_2O[/itex] reacts with 125.0 mL of the 0.1717 M basic solution of [itex]Na_4XeO_6[/itex].

I think now my corrected answer is  correct.

Title: Re: Occurrences, preparations and properties of Noble gases
Post by: Borek on April 09, 2023, 04:31:02 AM

Looks OK.