Which chemical bonds are found within water molecules?
1 answer:
You might be interested in
The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas
The reaction is second order for with a rate constant of
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
where,
k = rate constant =
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
= Initial concentration = 0.260 M
Putting values in above equation, we get:
Hence, the time taken is 5.19 seconds
Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al: 2 moles
- H₂SO₄: 3 moles
- Al₂(SO₄)₃. 1 mole
- H₂: 3 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- H₂SO₄: 98 g/mole
- Al₂(SO₄)₃: 342 g/mole
- H₂: 2 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 2 moles ×27 g/mole= 54 grams
- H₂SO₄: 3 moles ×98 g/mole= 294 grams
- Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
- H₂: 3 moles ×2 g/mole= 6 grams
The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?
<u><em>mass of aluminium sulfate= 88.67 grams</em></u>
Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.
<h3>Percent yield</h3>The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
<h3>Percent yield for the reaction in this case</h3>In this case, you know:
- actual yield= 75.26 grams
- theorical yield= 88.67 grams
Replacing in the definition of percent yields:
Solving:
<u><em>percent yield= 84.88%</em></u>
Finally, the percent yield for the reaction is 84.88%.
Learn more about
the reaction stoichiometry:
percent yield:
#SPJ1