Calculate the volume of O2 at STP that will be formed by the thermal decomposition of 24.5g
potassium chlorate.
Answers
Answer:
The formula for the decomposition reaction of potassium chlorate is
2KClO3→2KCl+3O2
That gives you 3 moles of oxygen for each 2 moles of potassium perchlorate.
The molecular weight of potassium perchlorate is 122.55 g/mol.
24.5 g is roughly 24.5122.55=0.2 mol
The number of mole of oxygen is therefore 0.3 mol.
At STP the volume can be approximated by using the ideal gas equation.
V=nRTP=0.3×8.314×273101.3≈6.8 litres
Answer:
It forms 6.7 lt of O₂
Explanation:
This is because,
The reaction of decomposition is;
2KClO₃→ 2KCl+3O₂
Therefore this indicates that every 2 moles of KClO₃ forms 3 moles of O₂.
We need to find how many moles is 24.5g of KClO₃ so that we can find the no. of moles O₂ forms through which we can further find the amount in grams.
No. of moles= amount given/molecular weight of the compound
= 24.5/122 (39+35+48)(molecular weight of KClO₃)
= 0.20 moles (approx)
Now,
2 moles of KClO₃₋₋₋₋₋₋ 3 moles of O₂
0.20 moles of KClO₃₋₋₋₋₋₋ x moles of O₂
x = 3×0.20/2 (cross multiplication)
= 0.3 moles
Now using ideal gas equation,
V= nRT/P
(where v= volume (cm³), n= no. of moles, R= ideal gas constant (8.314 J/mol·K), T= temperature (K), P= pressure (pa)
V= 0.3×8.314×273/101325
(since we are taking at STP, T will be 273K and pressure is 1 atm i.e, when converted to pascal is 101325pa)
∴ V= 0.0067 cm³ (approx)
Since 1 cm³ = 1ml
V= 0.0067 ml
= 6.7 lt