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CaHeK987 [17]
2 years ago
13

How many molecules of water are in 0.578 moles of water?

Chemistry
1 answer:
Elena L [17]2 years ago
3 0
1 mole contains 6.02 * 10^23 molecules
0.578 moles of water contains 0.578 * 6.02 * 10^23 molecules
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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
Rufina [12.5K]

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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Explanation:

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Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
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Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

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