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Evgen [1.6K]
3 years ago
12

How many moles are there in 78.3g of CO2?

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
4 0
The molar mass of CO2 can be calculated as follows;
CO2 — 12 + (16x2) = 12+ 32 = 44 g
Therefore molar mass of CO2 is 44 g/mol
In 44 g of CO2 there’s 1 mol of CO2
Then 1 g of CO2 there’s 1/44 mol of CO2
Therefore in 78.3 g of CO2 there’s — 1/44 x 78.3 =1.78 mol of CO2
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You are on an alien planet where the names for substances and the units of measures are very unfamiliar. Nonetheless, you obtain
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Answer: 12


Explanation.


This problem was recently posted and deals with the conversion of units,so I give the same answer.


You need to convert the 9 quibs of skvarnick to sleps, to determine how many sleps you have.


The conversion of units is easily undertaken once you build your conversion factors.


Conversion factors are ratios (fractions) equivalent to one, which permits use the identity property of the multiplication to change one unit to a different equivalent one.


Remember that the identity property of multiplication states that any amount multiplied by 1 remains unchanged, i.e.

               a.1=a

Then, build your conversion factor from the definitions given:

  •    12 sleps is equal to 9 quibs.
  •    12 sleps = 9 quibs
  •    in virtue of the division property of equality, you can divide both sides by 12 sleps and get:

                                       1 = 12 sleps /  9 quibs ← conversion factor

Now you can multiply the amount 9 quibs by the conversion factor, which is equal to 1, to find the equivalent amount of sleps:


  •    identity property: 9 quibs = 9 quibs × 1
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You can see that the unit quibs appear in the numerator and the denominator so it cancels out, and the result will be in sleps.


So, all you have to do now is the operations with the numbers:

       9 quibs = (9 ×  12 / 9) sleps = 12 sleps ← answer

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Answer: Option (c) is the correct answer.

Explanation:

A limiting reagent is defined as a reagent that completely gets consumed in a chemical reaction. A limiting reagent limits the formation of products.

For example, we have given 5 mol of A and the reaction is 2A + 4B \rightarrow 2AB

Whereas when 4 mol B will react with 2 mol of A. Hence, 8 mol of B will react with 4 mol A as follows.

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As, the given moles of A is more than the required moles. Thus, it is considered as an excess reagent.

Hence, B is a limiting reagent because it limits the formation of products.

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