What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in the reaction $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ ?

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What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in the reaction $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ ?

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Answer 1 · 2017-03-12T00:08:01

$1 : 1$ $1:1$

Explanation:

The mole ratio that exists between two chemical species that take part in a chemical reaction is simply the ratio that exists between the stoichiometric coefficients added in front of said species.

In your case, you have

$2 {CO}_{(g)} + O_{2 (g)} \to 2 {CO}_{2 (g)}$ $color(blue)(2)"CO"_ ((g)) + "O"_ (2(g)) -> color(red)(2)"CO"_ (2(g))$

According to the balanced chemical equation that describes this reaction, the reaction consumes $2$ $color(blue)(2)$ moles of carbon monoxide and $1$ $1$ mole of oxygen gas and produces $2$ $color(red)(2)$ moles of carbon dioxide.

So, carbon monoxide has a coefficient of $2$ $color(blue)(2)$ . The same can be said about carbon dioxide, which has a coefficient of $2$ $color(red)(2)$ as well.

This means that the mole ratio that exists between carbon monoxide and carbon dioxide will be

$\frac{2 . moles CO}{2 . {moles CO}_{2}} = \frac{2}{2} = \frac{1}{1}$ $(color(blue)(2)color(white)(.)"moles CO")/(color(red)(2)color(white)(.)"moles CO"_2) = color(blue)(2)/color(red)(2) = 1/1$

Because the two coefficients are equal, we say that the two compounds have a $1 : 1$ $1:1$ mole ratio.

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What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in the reaction $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ ?

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What is the mole ratio of CO(g)CO(g) to CO2(g)CO_2(g) in the reaction 2CO(g)+O2(g)→2CO2(g)2CO(g) + O_2(g) -> 2CO_2(g)?

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What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in the reaction $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ ?