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
Answer:
44
Explanation:
protons and electrons have the same number:)
Balance the equation first:
2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)
Moles of KClO3 = 110 / 122.5 = 0.89
Following the balanced chemical equation:
We can say moles of O2 produce =
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x moles of KClO3
So, O2 = (3 / 2) x 0.89
= 1.34 moles
So, Volume at STP = nRT / P
T = <span>273.15 K
P = 1 atm
So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>
Answer:
Explanation:
The Ideal Gas Law states that PV=nRT.
Rearrange that into P/n=RT/V.
In this case, the cylinder is rigid so the volume, V, does not change.
Temperature does not change either.
Out of 450 grams of gas, 150 grams leak out. So only 450-150 = 300 grams is left.
n is number of moles which is dependent on mass:
n1/n2 = 450/300 = 3/2
P1/n1 = RT/V = P2/n2
P2 = P1/n1*n2
= 7.2/3*2
= 4.8 atmosphere