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# What is the value of $\Delta {n_g}$ if we consider the combustion of $1mol$ of liquid ethanol if reactants and products are at $298K$:A. $- 1$ B. $2$ C. $+ 1$D. $+ 2$

Last updated date: 20th Jun 2024
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Hint: The regulation of preservation of mass or principle of mass conservation defines that for any system which is closed to all transfers of count number and energy, then the mass of the device have to remain steady through the years, because the machine's mass cannot exchange, so quantity can neither be added nor be removed.

Complete step by step solution:
Giving the law of conservation of mass, mass can neither be created nor be destroyed. Therefore the mass of products has to be equal to the mass of reactants. The quantity of atoms of every detail needs to be identical on reactant and product aspect. Thus chemical equations are balanced.
The balanced equation for combustion of ethanol at $298K$ is
${C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l)$
$\Delta {n_g}$ is the change in number of moles of gas particles given below:
${n_{products}} - {n_{reactants}} = 2 - 3 = - 1$
So , the value of $\Delta {n_g}$ is if we consider the burning of one mol of liquid ethanol if reactants and products are at $298K$ is $- 1$.

So, the correct answer is A.

The Law of Conservation of Mass periods from Antoine Lavoisier's $1789$ detection that mass is neither produced nor ruined in the chemical reactions. In different phrases, the mass of anyone element at the start of a response will be the same as the mass of that element at the give up of the response.