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In a combustion reaction, 46.0g of ethanol reacts with 96.0 g of oxygen to produce water and carbon dioxide. If 54.0 g of water is produced, what mass of carbon dioxide is produced?

User GeckoTang
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Final answer:

The mass of carbon dioxide produced in the combustion reaction of ethanol with oxygen, given that 54.0 g of water is produced, is 88.0 g.

Step-by-step explanation:

In a combustion reaction of ethanol (C₂H₅OH), which reacts with oxygen (O₂) to produce water (H₂O) and carbon dioxide (CO₂), the mass of carbon dioxide produced can be calculated using the law of conservation of mass. The balanced chemical equation for the combustion of ethanol is:

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

Given 46.0 g of ethanol reacts with 96.0 g of oxygen, and 54.0 g of water is produced, we can deduce the mass of carbon dioxide. By the conservation of mass, the total mass of reactants must be equal to the total mass of products:

Mass of reactants = Mass of products

46.0 g (ethanol) + 96.0 g (oxygen) = Mass of water + Mass of carbon dioxide

142.0 g (total reactants) = 54.0 g (water) + Mass of carbon dioxide

Mass of carbon dioxide = 142.0 g - 54.0 g = 88.0 g of CO₂

Therefore, the mass of carbon dioxide produced in this reaction is 88.0 g.

User Learn AspNet
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The basic law of mass conservation states that, the total mass of products of a chemical reaction should be equal to the total mass of reactants in the same reaction.

Applying this law to the above problem, we find that:
Total mass of reactants = 46 + 96 = 142 grams
This means that the total mass of products (water + carbon dioxide) should be also 142 grams
Since 52 grams of water is produced, therefore,
mass of carbon dioxide = 142 - 52 = 90 grams
User IDDQD
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