Given the balanced equation representing a reaction: $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ . What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in this reaction?

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Given the balanced equation representing a reaction: $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ . What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in this reaction?

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Answer 1 · 2018-05-20T18:11:20

Is it not $1 : 1$ $1:1$ ...?

Explanation:

We could write... $C O (g) + \frac{1}{2} O_{2} \to C O_{2} (g)$ $CO(g) +1/2O_2 rarr CO_2(g)$ , instead of ...

$2 C O (g) + O_{2} \to 2 C O_{2} (g)$ $2CO(g) +O_2 rarr 2CO_2(g)$

In either scenario both mass and charge are balanced, as required for a stoichiometric equation...and this is why teachers go to such great efforts to teach stoichiometry....and we could put in numbers to stress the mass balance of the equation....

$underbrace(CO(g) +1/2O_2)_"28 g + 16 g = 44 g" rarr underbrace(CO_2(g))_"44 g"$

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Given the balanced equation representing a reaction: $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ . What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in this reaction?

1 Answer

Explanation:

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Given the balanced equation representing a reaction: 2CO(g) + O_2(g) -> 2CO_2(g)2CO(g)+O2(g)→2CO2(g)2CO(g) + O_2(g) -> 2CO_2(g). What is the mole ratio of CO(g)CO(g)CO(g) to CO_2(g)CO2(g)CO_2(g) in this reaction?

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Explanation:

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Given the balanced equation representing a reaction: $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) -> 2CO_2(g)$ . What is the mole ratio of $C O (g)$ $CO(g)$ to $C O_{2} (g)$ $CO_2(g)$ in this reaction?