How many moles of oxygen are needed for the complete combustion of 29.2 grams of acetylene?

Moles of Oxygen= 2.8075 moles

Given

29.2 grams of acetylene

Required

moles of Oxygen

Solution

Reaction(Combustion of Acetylene) :

2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)

Mol of Acetylene :

= mass : MW Acetylene

= 29.2 g : 26 g/mol

= 1.123

From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :

= 5/2 x mol C₂H₂

= 5/2 x 1.123

= 2.8075 moles

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consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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