Question

A $560\;ml$ of carbon monoxide is mixed with $500\;ml$ of oxygen and ignited. The chemical equation for the reaction is as follows:
$2CO + {O_2} \to 2C{O_2}$
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
A. $200ml,600ml$
B. $280ml,560ml$
C. $560ml,280ml$
D. $500ml,200ml$

Answer 1

Hint: As we know that the stoichiometric coefficient in a chemical reaction basically helps us in identifying the number of moles or molecules or atoms or volume of any substances which can react with another substance to form a particular product.

Complete step by step solution:
As we know that the stoichiometric coefficients are used in calculating moles, molecules, atoms or even volumes of any substances. Similarly, in the equation given below:
$2CO + {O_2} \to 2C{O_2}$

- We can say that two moles of carbon monoxide reacts with one mole of oxygen to give two moles of carbon dioxide or we can say that two molecules of carbon monoxide reacts with one molecule of oxygen to form two molecules of carbon dioxide.
- In terms of volume, we can say that to burn two volumes of carbon monoxide, one volume of oxygen is required and it results in the production of two volumes of carbon dioxide.
- Similarly, to burn $560\;ml$ of carbon monoxide, volume of oxygen required will be equivalent to $\dfrac{{560}}{2} = 280ml$ and therefore, $560\;ml$ of carbon dioxide will be produced.
- Therefore, the volume of oxygen used in the above chemical reaction is $280ml$ and the volume of carbon dioxide formed in the reaction is $560\;ml$.

Therefore, the correct answer is (B).

Note: Always remember that the mole concept is the easiest way to calculate the amount of any reactant as well as any product. Also, the mole of any substance is given by the ratio of its mass to the molecular mass which in turn is equivalent to the ratio of volume in litres at STP to $22.4L$ and this ratio can be put equal to the ratio of number of atoms or molecules to the Avogadro’s number. Hence, any parameter can be calculated using these ratios which can be given as:
$moles = \dfrac{{mass}}{{mol.\;mass}} = \dfrac{{Vol.\;in\;L}}{{22.4L}} = \dfrac{{no.of\;atoms\;or\;molecules}}{{{N_A}}}$