Which element would be a strong reducing agent? F2 Ba Na Cl2

Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32

4 0

3 years ago

For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i

frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
The charge on simple ions signifies their oxidation number.
The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

K + (-1) + 2(-2) = 0

K-5 = 0

K = +5

The oxidation number of K in KCl:

K + (-1) = 0

K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0

3 years ago

Read 2 more answers

A sample of aluminum is placed in a 25-ml graduated cylinder containing 10.0 ml of water. the level of the water rises to 18.0 m

OLga [1]

To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.

Then knowing volume of aluminum and the density of it, we can solve for the mass.

D = m/v

Dv = m

2.7 g/ml • 8 ml = 21.6 grams.

5 0

3 years ago

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if 15 mol C is mixed with 10 mol O2 which reactant is the limiting reactant? Which reactant would be the excess reactant?

ioda

Answer:

C is the excess reactant.

Explanation:

Reaction is C + O2 --> CO2

1mol of C required to react with 1mol O2

Therefore 15 - 10 = 5moles of C will be in excess

6 0