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1 answer:
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Answer:
1.635 M
Explanation:
Given:
10 mL of 20 volumes Hydrogen Peroxide
Here,
20 volumes of Hydrogen Peroxide means that on decomposition of 1 mL of H₂O₂ 20 mL of O₂ is obtained
also,
means 1 dm³ of H₂O₂ solution produces 20 dm³ oxygen
Now,
at 298K and 1 atm
20 dm³ oxygen = moles
or
= 0.817 moles
also,
2H₂O₂ → 2H₂O + O₂
thus,
1 dm³ of solution must contain 2 × moles of O₂ as moles of H₂O₂
thus,
Number of moles of H₂O₂ = 2 × 0.817
or
Number of moles of H₂O₂ = 1.635 moles
Hence,
For 20 volume hydrogen peroxide is 1.635 M
This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:
Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:
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