Question

The drug nicotine has the molecular formula ${C_{10}}{H_{14}}{N_2}.$ If 0.1 mole of this is combusted, what would be the weight of $C{O_2}$ obtained?
A. $ 440\theta $
B.$ 4.4\theta $
C.$ 44\theta $
D.$ 100g$

Answer 1

Hint:Combustion, or burning, is a high-temperature exothermic redox chemical reaction between a fuel (the reductant) and an oxidant, usually atmospheric oxygen, that produces oxidized, often gaseous products, in a mixture termed as smoke. A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve ${O_2}$ as one reactant.

Complete step by step answer:
The combustion of hydrogen gas produces water vapor.
$2{H_2}\left( g \right) + {O_2}\left( g \right) \to 2{H_2}O\left( g \right)$
Here,
Nicotine $\left( {{C_{10}}{H_{14}}{N_2}} \right)$ and oxygen are the reactants.
Now, write the Skeleton equation.
${C_{10}}{H_{14}}{N_2} + {O_2} \to C{O_2} + {H_2}O + {N_2}$
Balance the equation
 $ {C_{10}}{H_{14}}{N_2} + 27/2\,{O_2} \to 10C{O_2} + 7{H_2}O + {N_2} \\$
$ 1\,mole\xrightarrow{{}}10\,moles \\$
$ 0.1\,mole\xrightarrow{{}}1\,mole \\ $
Now, the weight of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\].
= Number of moles × M
 \[ = 1 \times 44\]
  \[ = 44\;{\rm{g}}\]

Note:
In complete combustion, the reactant burns in oxygen and produces a limited number of products. When a hydrocarbon burns in oxygen, the reaction will primarily yield carbon dioxide and water.
Incomplete combustion will occur when there is not enough oxygen to allow the fuel to react completely to produce carbon dioxide and water. It also happens when the combustion is quenched by a heat sink, such as a solid surface or flame trap. As is the case with complete combustion, water is produced by incomplete combustion; however, carbon, carbon monoxide, and hydroxide are produced instead of carbon dioxide.