Question

How do mole ratios compare to volume ratios ?

Answer 1

The molar ratio of the two gases is created from the coefficients in the balanced equation. Since the molar ratio is the same to the volume ratio for gases at the same temperature and pressure, the coefficients after the balanced equation can similarly be used to generate a influence that converts straight from volume of one gas to volume of another, as in the following sample
$N{H_3}\left( g \right){\text{ }} + {\text{ }}2{O_2}\left( g \right)\;\; \to \;\;HN{O_3}\left( l \right){\text{ }} + {\text{ }}{H_2}O\left( l \right)$

Complete step by step answer:
Let us assume that we are discussing reactions between ideal gases, for which a relationship between molar ratio and volume ratio can be simply determined.
So, affording to Avogadro's law, equal volumes of ideal gases comprise the same number of moles at the similar temperature and pressure. These resources that if pressure and temperature are detained constant, 1 mole of any ideal gas will occupy the precise similar volume.
${V_{molar}} = {\text{ }}V/n{\text{ }}\;for\;n = 1\;mole$........(i)
The most communal submission of this principle is the molar volume of a gas at STP - Standard Temperature and Pressure. At STP conditions, which suggest a temperature of $273.15\;K\;$ and a pressure of $1\;atm$, $1\;mole$ of any ideal gas occupies a volume of accurately $22.4\;L$.
So, let's say we have a reaction that involves ideal gases at STP.
${N_2}\left( g \right) + 3{H_2}\left( g \right){\text{ }} \to {\text{ }}2N{H_3}\left( g \right)$
The molar ratio between ${N_2},{H_2},N{H_3}\;$ is $1:3:2$; that is, $1\;mole$ of ${N_2}$ ​ reacts with $3\;moles$ of ${H_2}\;$ to produce $2\;moles$ of $N{H_3}$ ​.
Let's take up $0.25\;moles$ of ${N_2}$​ that react totally with ${H_2}\;$​ to produce $N{H_3}$​. Since we are STP, we recognize that $1\;mole$ of every of these gases occupies $22.4\;L$. Conferring to equation (i), this means that:
$V{\text{ }}{H_2}{\text{ }} = {\text{ }}n{\text{ }}{H_2} \times {\text{ }}{V_{molar}} = {\text{ }}3{\text{ }} \times {\text{ }}0.25{\text{ }} \times {\text{ }}22.4\;L{\text{ }} = {\text{ }}16.8\;L$
$V{\text{ }}{N_2} = {\text{ }}n{\text{ }}{N_2} \times {\text{ }}{V_{molar}} = {\text{ }}0.25{\text{ }} \times {\text{ }}22.4\;L{\text{ }} = {\text{ }}5.6\;L$
$V{\text{ }}N{H_3}{\text{ }} = {\text{ }}n{\text{ }}N{H_3}{\text{ }} \times {\text{ }}{V_{molar}}{\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}0.25{\text{ }} \times {\text{ }}22.4\;L{\text{ }} = {\text{ }}11.2\;L$
The volume ratio between ${N_2}$, ${H_2}\;$​, and $N{H_3}$ will be $5.6{\text{ }}:16.8{\text{ }}:11.2$, which is, obviously, equal to$\;1{\text{ }}:{\text{ }}3{\text{ }}:{\text{ }}2$ , the molar ratio between the gases.

Note:
So, for gaseous reactants and products that are at similar temperature and pressure, the molar ratio is equal to the volume ratio.