Question

A certain mass of carbon required $\,16\,grams\,$ of oxygen to be converted into carbon monoxide. To convert $\,CO\,$ into carbon dioxide $\,(C{O_2})\,$, the mass of oxygen required would be:
A.$\,4.0grams\,$
B.$\,8.0grams\,$
C.$12grams\,$
D.$\,16grams\,$

Answer 1

In order to find the answer to this question, we have to first find the number of moles of all the elements in each reaction. Here, we have two reactions involved, one is the formation of carbon monoxide and the other is the formation of carbon dioxide.
Molecular mass of the compounds;
$\,C = 12g/mol\,$
$\,{O_2} = 32g/mol\,$
$\,CO = 28g/mol\,$
$\,C{O_2} = 44g/mol\,$

Complete step by step answer:
Let us first look into the balanced chemical reaction for the formation of carbon monoxide;
$\,2C + {O_2} \to 2CO\,$
Here, we have to calculate the number of moles for each element;
$\,Number\,of\,moles\,of\,C = 2\,$
$\,Number\,of\,moles\,of\,{O_2} = 1\,$
$\,Number\,of\,moles\,of\,CO = 2\,$
Number of moles of each element has been found out from the chemical reaction.
But, in the question it is given that only $\,16g\,$ of oxygen was required in the specified reaction.
Hence, the number of moles will be reduced to half in the specified reaction mentioned in the question; the number of moles for carbon, oxygen and carbon monoxide would be $\,1,0.5\,$ and $\,1\,$respectively. So, here the molar masses are also reduced to half
Now, let us look into the reaction for the formation of carbon dioxide;
$\,2CO + {O_2} \to 2C{O_2}\,$

Number of moles of elements in the original reaction of formation of $\,C{O_2}\,$	$\,2\,$	$\,1\,$	$\,2\,$
Number of moles of elements in the reaction of formation of $\,\,C{O_2}\,$ mentioned in the question (reduced to half)	$\,1\,$	$\,0.5\,$	$\,1\,$

However, here the amount carbon monoxide participated is equal to that which produced in the above reaction. As only half of the carbon monoxide is produced in the reaction, the number of moles is reduced to half to all the elements here also. Hence, here also only $\,0.5\,$ moles of oxygen are being participated and hence the mass of that oxygen can be calculated by;
$\,Number\,of\,moles = \dfrac{{Given\,mass}}{{Gram\,molecular\,mass}}\,$
$\,\therefore Given\,mass = \dfrac{{Number\,of\,moles}}{{Gram\,molecular\,mass}}\,$
$\, = 32 \times 0.5\,$
$\, = 16grams\,$
Therefore, the answer to the above question is $\,16grams\,$ and hence, Option D is the correct answer

Note:
The common reaction mentioned in the solution is not the only reaction in which carbon dioxide can form. It occurs from the oxidation of any material containing carbon (wood, methane, propane, fuel, etc.), as well as from the action of the acid on the carbonate. However, in carbon dioxide, carbon and oxygen are present in the ratio $\,1:2\,$.