How many liters of O2 gas will be produced at STP from 4 moles of KC1O3

Chemistry

1 answer:

quester [9]3 years ago

5 0

Answer:

V = 134.5 L

Explanation:

Given data:

Number of moles of KClO₃ = 4 mol

Litters of oxygen produced at STP = ?

Solution:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

Now we will compare the moles of KClO₃ with oxygen.

KClO₃ : O₂

2 : 3

4 ; 3/2×4 = 6 mol

Litters of oxygen at STP:

PV = nRT

V = nRT/P

V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm

V = 134.5 L / 1

V = 134.5 L

You might be interested in

Explain why the ionization energy to remove a second electron from potassium is higher than the ionization energy to remove four

never [62]

Explanation:

It is more difficult to remove electrons from the second shell or energy level because of the imbalance between the positive nuclear charge and the remaining electrons.

The amount of energy required to remove electrons in ground state of an atom is the ionization energy.
The first ionization energy is the energy needed to remove the most loosely bound electron of an atom in the gas phase in ground state.
The second energy has a greater nuclear pull as it is closer to the nucleus.
Both potassium and silicon have the same number of energy levels.

6 0

3 years ago

The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of ene

sesenic [268]

Answer:

$T_2=98.5^{\circ}$

Explanation:

Given that,

The specific heat of water is 4.184Jg°C

Mass, m = 600 g

Initial temperature, T₁ = 75.5°C

We need to find the final temperature. We know that heat absorbed is given by :

$Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}$