400cm³ of dry air was passed over heated copper powder in a hard glass tube until no further change. What volume of remained at
the end of the experiment?
2 answers:
Answer:320cm³
Explanation:
20% of oxygen is present in 100% of air
xcm³ of O2 is present in 400cm³ of air
Therefore
80cm³ of O2 is present in 400cm³ of air
Vol of air remained =400-80
=320cm³ of air
2Cu+O2=>2CuO
2mol. 1mol. 2mol
160cm³. 80cm³ 160cm³
Since O2 is used up in the reaction
Vol of air remained=320cm³
Mark please
Explanation:
I think there' will be a decrease in volume of the air.
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Mass of KCl= 19.57 g
<h3>Further explanation</h3>
Given
12.6 g of Oxygen
Required
mass of KCl
Solution
Reaction
2KClO3 ⇒ 2KCl + 3O2
mol O2 :
= mass : MW
= 12.6 : 32 g/mol
= 0.39375
From the equation, mol KCl :
= 2/3 x mol O2
= 2/3 x 0.39375
=0.2625
Mass KCl :
= mol x MW
= 0.2625 x 74,5513 g/mol
= 19.57 g
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