Answer:
0.01918 moles of O₂
1.30g of H₂O₂
Explanation:
The H₂O₂ is descomposed to O₂ in the presence of MnO₂
2 H₂O₂(aq) → O₂(g) + 2 H₂O(l)
In beaker 1, there is no reaction, and mass of H₂O₂ is:
0.0200L× (9.77mol / L) = 0.1954mol × (34g / mol) = 6.64g. And mass of beakers is:
43.09g - 6.64g = 36.45g
In the second beaker, mass of reaction is:
42.58g - 36.45g -0.1g of MnO₂ =
6.03g = 34X + 18Y (1)
Where X are moles of H₂O₂ and Y moles of water
Also, moles of H₂O₂ and H₂O are:
0.1954mol = X + Y (2)
Replacing (1) in (2):
6.03g = 34(0.1954-Y) + 18Y
6.03 = 6.6436 -34Y + 18Y
-0.6136 = -16Y
0.03835 = Y = moles of water
Based on the reaction, 2 moles of water are produced whereas 1 mol of Oxygen is produced. Thus, moles of oxygen are:
0.03835mol H₂O × (1mol O₂ / 2mol H₂O) = 0.01918 moles of O₂
Moles produced of H₂O are the same than moles of H₂O₂ descomposed. Thus, mass of H₂O₂ descomposed is:
0.03835mol ₓ (34g / mol) = 1.30g of H₂O₂
In this exercise we have to use the knowledge of moles to calculate the amount of mass that will be produced, in this way we find that:
0.01918 moles of O₂ in 1.30g of H₂O₂
So knowing that the decomposition equation will be formed by:
[tex]2 H_2O_2(aq) \rightarrow O_2(g) + 2 H_2O(l)[/tex]
So calculating the amount of H2O2 in the reaction, we find that:
[tex]0.0200L* (9.77mol / L) = 0.1954mol * (34g / mol) = 6.64g\\ 43.09g - 6.64g = 36.45g [/tex]
In the second beaker, mass of reaction is:
[tex]42.58g - 36.45g -0.1g \\ =6.03g = 34X + 18Y (1) [/tex]
Creating a system with the values found earlier, we have:
[tex]0.1954mol = X + Y (2)\\ 6.03g = 34(0.1954-Y) + 18Y\\ 6.03 = 6.6436 -34Y + 18Y\\ -0.6136 = -16Y\\ 0.03835 = Y [/tex]
See more about moles at brainly.com/question/15209553
