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Topic: Occurrences, preparations and properties of Noble gases  (Read 2106 times)

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Offline Win,odd Dhamnekar

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Occurrences, preparations and properties of Noble gases
« 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?
« Last Edit: April 08, 2023, 05:41:05 AM by Win,odd Dhamnekar »
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Offline Hunter2

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Re: Occurrences, preparations and properties of Noble gases
« Reply #1 on: April 08, 2023, 07:01:24 AM »
It start the equation is wrong, not balanced

Error with the brackets at Manganese nitrate.

Offline Win,odd Dhamnekar

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Re: Occurrences, preparations and properties of Noble gases
« Reply #2 on: April 08, 2023, 09:36:03 AM »
It start the equation is wrong, not balanced

Error with the brackets at Manganese nitrate.

Then what is the correct balanced chemical equation?
Any science consists of the following process.
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5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline Hunter2

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Re: Occurrences, preparations and properties of Noble gases
« Reply #3 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.

Offline billnotgatez

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Re: Occurrences, preparations and properties of Noble gases
« Reply #4 on: April 08, 2023, 10:29:59 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

Offline Win,odd Dhamnekar

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Re: Occurrences, preparations and properties of Noble gases
« Reply #5 on: April 09, 2023, 02:36:45 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.

Any science consists of the following process.
 1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline Borek

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Re: Occurrences, preparations and properties of Noble gases
« Reply #6 on: April 09, 2023, 04:31:02 AM »
Looks OK.
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