Two samples of hot unknown metals of equal temperatures are dropped in separate containers of water at 30 degrees Celsius. The f
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<u>Answer:</u>
<u>For a:</u> The number of molecules of nitrogen dioxide is
<u>For b:</u> The mass of nitric acid formed is 54.81 grams
<u>For c:</u> The mass of nitric acid formed is 206 grams
<u>Explanation:</u>
The given chemical reaction follows:
- <u>For a:</u>
By Stoichiometry of the reaction:
1 mole of water reacts with 3 moles of nitrogen dioxide
So, 0.250 moles of water will react with of nitrogen dioxide
According to mole concept:
1 mole of a compound contains number of molecules.
So, 0.75 moles of nitrogen dioxide will contain number of molecules
Hence, the number of molecules of nitrogen dioxide is
- <u>For b:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of nitrogen dioxide = 60.0 g
Molar mass of nitrogen dioxide = 46 g/mol
Putting values in equation 1, we get:
By Stoichiometry of the reaction:
3 moles of nitrogen dioxide produces 2 mole of nitric acid
So, 1.304 moles of nitrogen dioxide will produce = moles of nitric acid
Now, calculating the mass of nitric acid from equation 1, we get:
Molar mass of nitric acid = 63 g/mol
Moles of nitric acid = 0.870 moles
Putting values in equation 1, we get:
Hence, the mass of nitric acid formed is 54.81 grams
- <u>For c:</u>
- <u>For nitrogen dioxide:</u>
Given mass of nitrogen dioxide = 225 g
Molar mass of nitrogen dioxide = 46 g/mol
Putting values in equation 1, we get:
- <u>For water:</u>
Given mass of water = 55.2 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
By Stoichiometry of the reaction:
3 moles of nitrogen dioxide reacts with 1 mole of water
So, 4.90 moles of nitrogen dioxide will react with = of water
As, given amount of water is more than the required amount. So, it is considered as an excess reagent.
Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 mole of nitrogen dioxide produces 2 moles of nitric acid
So, 4.90 moles of nitrogen dioxide will produce of nitric acid
Now, calculating the mass of nitric acid from equation 1, we get:
Molar mass of nitric acid = 63 g/mol
Moles of nitric acid = 3.27 moles
Putting values in equation 1, we get:
Hence, the mass of nitric acid formed is 206 grams