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GarryVolchara [31]
3 years ago
7

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): 2H2O2

(aq) ---> 2H2O(l) + O2(g)
What volume of pure O2(g), collected at 27C and 746 torr, would be generated by decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution?
Chemistry
1 answer:
zalisa [80]3 years ago
8 0

Answer:

23.0733 L

Explanation:

The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

Mass=\frac {50}{100}\times 125\ g

Mass = 62.5 g

Molar mass of H_2O_2 = 34 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus, moles are:

moles= \frac{62.5\ g}{34\ g/mol}

moles= 1.8382\ mol

Consider the given reaction as:

2H_2O_2_{(aq)}\rightarrow2H_2O_{(l)}+O_2_{(g)}

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.

Also,

1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.

So,

1.8382 moles of hydrogen peroxide decomposes to give \frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr
The conversion of P(torr) to P(atm) is shown below:
[tex]P(torr)=\frac {1}{760}\times P(atm)

So,

Pressure = 746 / 760 atm = 0.9816 atm

Temperature = 27 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (27 + 273.15) K = 300.15 K

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K

<u>⇒V = 23.0733 L</u>

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How many moles of NH3 would be formed from the complete reaction of 16.0 g H2?
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Taking into account the reaction stoichiometry, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

N₂ + 3 H₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • N₂: 1 mole
  • H₂: 3 moles
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The molar mass of the compounds is:

  • N₂: 14 g/mole
  • H₂: 2 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • N₂: 1 mole ×14 g/mole= 14 grams
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<h3>Mass of NH₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 2 moles of NH₃, 16 grams of H₂ form how many moles of NH₃?

moles of NH_{3}= \frac{16 grams of H_{2} x2moles of NH_{3}}{6 grams of H_{2}}

<u><em>moles of NH₃= 5.33 moles</em></u>

Then, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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