Question

If $ 4.95g $ of Ethylene is combusted with $ 3.25g $ of Oxygen.
a) What is the limiting reagent?
b) How many grams of $ C{O_2} $ are formed?

Answer 1

Hint: Ethylene is the common name for ethane and is an unsaturated carbon compound consisting of two carbon atoms. The combustion reaction involves the burning of ethylene in the presence of oxygen to give out carbon dioxide and water as the products.

Complete answer:
The combustion reaction of hydrocarbons involves the reaction between the hydrocarbon and oxygen that results in the formation of carbon dioxide and water. The balanced chemical equation for the combustion of ethylene can be written as follows:
 $ {C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O $
The stoichiometric coefficients suggest that one mole of ethylene requires three moles of oxygen to produce two moles of carbon dioxide gas and two moles of water vapour.
Thus the ratio in which ethylene and oxygen react is $ 1:3 $
The number of moles of ethylene and oxygen can be calculated by finding out the ratio of their mass and molar mass.
 $ n = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}} $
Number of moles of ethylene are:
 $ n({C_2}{H_4}) = \dfrac{{4.95g}}{{28gmo{l^{ - 1}}}} = 0.177 $
Number of moles of oxygen are:
 $ n({O_2}) = \dfrac{{3.25g}}{{32gmo{l^{ - 1}}}} = 0.101 $
The limiting reagent is a reactant that is present in a lesser amount and gets consumed early in the reaction.
To maintain a $ 1:3 $ ratio between ethylene and oxygen, $ 0.177 \times 3 $= 0.531g of oxygen is needed but a lesser amount of oxygen is available for the reaction. Therefore, oxygen is the limiting reagent.
The reaction will proceed according to the amount of limiting reagents as once it gets consumed, the reaction stops. Thus, for each mole of oxygen a two-third $ \left( {\dfrac{2}{3}} \right) $ mole of carbon dioxide is formed as the ratio of oxygen reacting and carbon dioxide being produced is $ 3:2 $ .
Therefore, the number of moles of carbon dioxide produced are two-third times the number of moles of oxygen reacted:
 $ n(C{O_2}) = 0.101 \times \left( {\dfrac{2}{3}} \right) = 0.0673 $
The moles can be used to calculate the mass of carbon dioxide produced by multiplying it with the molar mass of carbon dioxide.
 $ n(C{O_2}) = 0.0673mol \times 44gmo{l^{ - 1}} = 2.9612g $
Hence, (a) The limiting reagent is oxygen and
(b) The mass of carbon dioxide produced is $2.9612g$.

Note:
The unsaturated compounds can be distinguished from the saturated compounds on the basis of the combustion reaction shown by them. The unsaturated compound burns with a sooty flame (black in colour) due to the presence of unburnt carbon particles in it.