Chemistry, 03.12.2019 18:31 fxllsxn

Determine the volume of o2 (at stp) formed when 50.0 g of kclo3 decomposes according to the following reaction. the molar mass for kclo3 is 122.55 g/mol. 22kclo3 (s) --> 2 kcl(s) + 3 o2(g)

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Taking into account the reaction stoichiometry, ideal gas law and the definition of STP conditions, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

The balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

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:

KClO₃: 2 moles KCl: 2 moles O₂: 3 moles

The molar mass for KClO₃ is 122.55 g/mol. Then, by reaction stoichiometry, the following mass quantity of the compound participates in the reaction:

$2 molesx\frac{122.55 g}{1 mole} =$ 245.1 grams of KClO₃

Then it is possible to apply the following rule of three: if by reaction stoichiometry 245.1 grams of KClO₃ form 3 moles of O₂, 50 grams of KClO₃ form how many moles of O₂?

$amount of moles of O_{2} =\frac{50 grams of KClO_{3} x3 moles of O_{2} }{245.1grams of KClO_{3}}$

amount of moles of O₂= 0.612 moles

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Then, in this case you know:

P= 1 atmV= ?n= 0.612 molesR= 0.082 $\frac{atmL}{molK}$ T= 0 C= 273 K

Replacing in the ideal gas:

1 atm× V= 0.612 moles× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$V=\frac{0.612 molesx 0.082\frac{atmL}{molK} x 273 K}{1 atm}$

V=13.70 L

Finally, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

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Answer from: Gearyjames8

STP conditions refer to the standard temperature and pressure. The volume of the oxygen at STP will be 13.70 L when 50 grams of potassium chlorate decomposes.

The chemical equation of decomposition if potassium chlorate can be written as:

$\text{KClO}_3\;\rightarrow\; 2\;\text{KCl + 3\; O}_2$

From the stoichiometry of the reaction, it can be observed as:

2 moles of potassium chlorate 2 moles of potassium chloride3 moles of oxygen

Now, calculating the mass of potassium chlorate participating in the reaction:

$2\;\text{moles}\;\times\dfrac{122.5\;\text g}{1 \;\text{mole}} &= 245.1 \;\text {g}\;\text {KClO}_3$

Now, substituting the values to calculate the moles of oxygen:

$\text{amount of moles oxygen}&=\dfrac{50\;\text g} {245.1 \;\text g} \times 3\;\text{moles of oxygen}$

Moles of oxygen = 0.612 moles

From the ideal gas equation:

PV = nRT, where,

P is pressure, Tis the temperature, and the volume, V. R is ideal gas constant.

Substituting the values:

1 atm x V = 0.612 moles x 0.082 x 273 KV = 13.70 L

Therefore, the 13.70 L of oxygen at STP will be formed when 50 grams of potassium chlorate decomposes into potassium and oxygen.

To know more about ideal gas equation, refer to the following link:

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Answer from: Quest

i believe it is a.) released into

explanation:

Answer from: Quest

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

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Determine the volume of o2 (at stp) formed when 50.0 g of kclo3 decomposes according to the followin...