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adelina 88 [10]
3 years ago
6

In virginia and some other parts of the world, water has carved vast caverns out of underground limestone deposits. water, howev

er, must contain dissolved carbon dioxide in order to break down the limestone. what role does the carbon dioxide play?
Chemistry
1 answer:
shusha [124]3 years ago
5 0
The role of the carbon dioxide in water is that it has the ability to form an acid in which it has the capacity to react  with underground limestones deposits. If it is able to react with the limestone, then it will most likely produce a substance that can be water-soluble.
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The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie
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<u>Answer:</u> Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}   ....(1)

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Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol

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Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol

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O_3+NO\rightarrow O_2+NO_2

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

  • Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

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