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

37.6 Km/h

Explanation:

From the question given above, the following data were obtained:

Distance = 100 m

Time = 9.58 s

Speed =?

Next, we shall convert 100 m to Km. This can be obtained as follow:

1000 m = 1 Km

Therefore,

100 m = 100 m × 1 Km / 1000 m

100 m = 0.1 km

Next, we shall convert 9.58 s to h. This can be obtained as follow:

3600 s = 1 h

Therefore,

9.58 s = 9.58 s × 1 h / 3600 s

9.58 s = 2.66×10¯³ h

Finally, we shall determine the speed. This can be obtained as follow:

Distance = 0.1 km

Time = 2.66×10¯³ h

Speed =?

Speed = Distance / Time

Speed = 0.1 / 2.66×10¯³

Speed = 37.6 Km/h