Explanation:
Starting moles of ethanol acid = 0.020 mol
At the equilibrium 50 % of the ethanol acid molecules reacted
∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %
= 0.010 mol
Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol
Moles of the product
gas formed are calculated as
0.010 mol CH3COOH * 1 mol
/ 2 mol CH3COOH
= 0.005 mol 
Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol 
That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Now Calculate the pressure :
0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas
P1/n1 = P2/n2
P2 = P1*n2 / n1
= 0.74 atm * 0.015 mol / 0.020 mol
= 0.555 atm
The illustration would be that of a double replacement reaction.
<h3>What are double replacement reactions?</h3>
They are reactions in which 2 ionic compounds exchange ions to form two new products.
Thus, in the reaction: ab + cd ----------> ad + cb
ab and cd are two ionic compounds. The b in ab is replaced by the d in cd while the d in cd itself is replaced by the b in ab. Hence, new products, ad and cd, are formed.
More on double replacement reactions can be found here: brainly.com/question/19267538
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I believe is different in pressure
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
