Equimolar amounts of Cl2(g) and CO(g) are injected into an evacuated, rigid container, where they react according to the equation below. Cl2(g)+CO(g)⇄COCl2(g) ΔHrxn=−109kJ/molrxn (a) If 7.0 g of CO(g) is consumed in the reaction with excess Cl2(g), how many moles of COCl2(g) are produced? (b) Which element is oxidized in this reaction? Justify your answer in terms of oxidation numbers.

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Answer:

0.25 moles of COCl₂ are been produced

The element that is oxidized is C, it changed the oxidation state from +2 in CO to +4 in phosgene.

Explanation:

Equilibrium reaction:

Cl₂(g)  +  CO(g)  ⇄  COCl₂(g)

Let's convert the mass of CO to moles:

7g . 1mol /28g = 0.25 moles

As ratio is 1:1, we can say that 0.25 moles of COCl₂ are been produced.

1 mol of chlorine reacts to 1 mol of carbon monoxide in order to produce 1 mol of phosgene.

Chlorine is been reduced:

Cl₂  +  2e⁻  ⇄  2Cl⁻

Change the oxidation state, from 0 (ground state) to -1. Oxidation state decreased.

Carbon is been oxidized.

In CO, carbon has +2 as oxidation state. In phosgene the oxidation state is +4. This oxidation state was increased, that's why it has oxidized.

The element that is oxidized is carbon whose oxidation number was increased from +2 to +4.

The equation of the reaction is;

Cl2(g) + CO(g) ⇄ COCl2(g)

We have been told that the chlorine gas is the reactant in excess hence the carbon monoxide is the limiting reactant.

Number of moles of  CO = 7.0 g/28 g/mol = 0.25 moles

Since the reaction is 1:1, 0.25 moles of COCl2 is produced.

The element that is oxidized is carbon whose oxidation number was increased from +2 to +4.

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