Question

When a mixture of 12.0 g of acetylene (c2h2 and 12.0 g of oxygen (o2 is ignited, the resultant combustion reaction produces co2 and h2o?

Answer 1

Number of moles of CO₂ formed = 0.375 moles

number of moles of water formed = 0.75 moles

Step-by-step 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 =
`(mass)/(molar mass) \\ = (12)/(26.04) \\= 0.461 moles`

Number of moles of O₂ =
`(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

Answer 2

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