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BaLLatris [955]
3 years ago
14

Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KClO3(s)2KCl(s) + 3O2(g) The prod

uct gas, O2, is collected over water at a temperature of 25 °C and a pressure of 749 mm Hg. If the wet O2 gas formed occupies a volume of 5.76 L, the number of moles of KClO3 reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

The number of moles of KClO₃ reacted was 0,15 mol

Explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)

V is volume (<em>5,76L</em>)

R is gas constant (<em>0,082 atmL/molK</em>)

And T is temperature (25°C ≡ <em>298,15K</em>)

Replacing, number of moles of O₂ are <em>0,2248 moles</em>

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂×\frac{2 mol KClO_{3}}{3 mol O_{2}} = <em>0,15 mol of KClO₃</em>

I hope it helps!

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Data Given:

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Solution:

               As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....

                          = ( 1 mole × 87.6 L) ÷ 22.4 L

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<h3>2nd Method:</h3>

                     Assuming that the gas is acting ideally, hence, applying ideal gas equation.

                              P V  =  n R T      ∴  R  =  0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹

Solving for n,

                              n  =  P V / R T

Putting values,

                              n  =  (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)

                              n  =  3.91 moles

Result:

          87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.

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