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Tju [1.3M]
3 years ago
15

When CO2 decomposes into oxygen and carbon, it gives a gram ratio of 2.67:1 O2:C. When a 32.4g of CO2 decomposes, how many grams

of carbon are produced, and how many grams of of O2 are produced? Hint(What is the total of carbon and oxygen produced)
Chemistry
1 answer:
Rufina [12.5K]3 years ago
6 0

Answer : The mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

CO_2\rightarrow O_2+C

As we are given:

\text{Mass of }O_2}:\text{Mass of }C=2.67:1{

According to the law of conservation of mass,

Total mass of CO_2 = Mass of O_2 + Mass of C

Total mass of CO_2 = 2.67 + 1 = 3.67 g

Now we have to calculate the mass of O_2 and C.

\text{Mass of }O_2=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }O_2=\frac{32.4g}{3.67g}\times 2.67=23.6g

and,

\text{Mass of }C=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }C=\frac{32.4g}{3.67g}\times 1=8.83g

Therefore, the mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

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1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

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The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

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If we replace this expression in (1) we have:

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