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

Answer 1

Is it not #1:1#...?

We could write...#CO(g) +1/2O_2 rarr CO_2(g)#, instead of ...

#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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Answer 2

Luna Chen

The mole ratio of CO(g) to CO2(g) in this reaction is 2:2, which simplifies to 1:1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Given the balanced equation representing a reaction: #2CO(g) + O_2(g) -&gt; 2CO_2(g)#. What is the mole ratio of #CO(g)# to #CO_2(g)# in this reaction?

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