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The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide. 
Balancing the combustion reaction,
                           C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
                        (12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
                         (12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
             CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
             H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g

tatendagota

Answer:

Number of moles of CO₂ formed = 0.375 moles

number of moles of water formed = 0.75 moles

Explanation:

Acetylene reacts with oxygen according to the following chemical reaction:

\(2C_{2}H_{2} + 5O_{2} = 4CO_{2} + 2H_{2} O\)

The molar ratios are :

C₂H₂ = 2

O₂     = 5

CO₂  = 4

H₂O  = 2

The basis is to find a limiting reagent which is given by the lowest number of moles.

The molar mass of acetylene is = 26.04

number of moles of acetylene   = \(\frac{mass}{molar mass} \\ = \frac{12}{26.04} \\= 0.461 moles\)

Number of moles of O₂ = \(\frac{12}{32} \\= 0.375 moles\)

Number of moles of CO₂ formed = 0.375 moles

number of moles of water formed (2 x 0.375) = 0.75 moles

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