Chemistry, 05.02.2020 11:50 abdullaketbi71

The depletion of ozone (o3) in the stratosphere has been a matter of great concern among scientists in recent years. it is believed that ozone can react with nitric oxide (no) that is discharged from the high-altitude jet plane, the sst. the reaction is
o3 + no > o2 + no2
if 0.827 g of o3 reacts with 0.635 g of no, how many grams of no2 will be produced? g no2 which compound is the limiting reagent? ozone (o3) nitric oxide (no) calculate the number of moles of the excess reagent remaining at the end of the reaction.

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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.

Explanation:

To calculate the number of moles, we use the equation

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

For ozone:

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

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

For nitric oxide:

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

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

For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

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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