Question

3g of carbon reacts completely with 5g of oxygen leaving no reactants. Find the mole fraction of carbon monoxide in the product gases.

Answer 1

Hint: To answer this question, you must recall the laws of stoichiometry, namely the law of multiple proportions, the law of definite proportions and the law of conservation of mass. Carbon on reaction with oxygen would give us a mixture of gaseous products containing both carbon monoxide and carbon dioxide.

Complete step by step solution:
From the law of multiple proportions, we can say that the reaction between carbon and oxygen occurs as follows:
 $ {\text{C}} + {{\text{O}}_2} \to {\text{C}}{{\text{O}}_2} + {\text{CO}} $
In the given question, we are given that 3g of carbon is reacted completely with 5g of oxygen. We are also given that none of the reactants are left, that means both the reactants react completely with each other. From the law of conservation of mass, we can say that all 3 grams of carbon and 5 grams of oxygen are converted into the products. Let the number of moles of carbon monoxide produced be $ x $ and the number of moles of carbon dioxide produced be $ y $ .
So we can write, $ x + y = \dfrac{3}{{12}} $ and $ x + 2y = \dfrac{5}{{16}} $
Solving these, we get $ x = \dfrac{3}{{16}} $ and $ y = \dfrac{1}{{16}} $
So, we can find the mole fraction of carbon monoxide as $ {\chi _{CO}} = \dfrac{{\dfrac{3}{{16}}}}{{\dfrac{3}{{16}} + \dfrac{1}{{16}}}} = \dfrac{3}{4} = 0.75 $
The mole fraction of carbon monoxide in the product mixture is $ 0.75 $ .

Note:
The law of multiple proportions proposes that two elements can combine in more than one proportion to give a variety of products. The law of definite proportions proposes that two elements for forming a certain compound must always react in a definite whole number ratio. The law of conservation of mass proposes that mass is neither created nor destroyed, or in other words, the mass of a system is always conserved.