[References) Liquid oxygen was first prepared by heating potassium chlorate, KCIO3, in a closed vessel to obtain oxygen at high pressure. The oxygen was cooled until it liquefied. 2KC103 (s) - 2KCI(s) + 302(9) If 176.0 g of potassium chlorate reacts in a 2.90-L vessel, which was initially evacuated, what pressure of oxygen will be attained when the temperature is finally cooled to 25°C? Use the preceding chemical equation, and ignore the volume of solid product. Pressure = atm O2
[References) Liquid oxygen was first prepared by heating potassium chlorate, KCIO3, in a closed vessel to obtain oxygen at high pressure. The oxygen was cooled until it liquefied. 2KC103 (s) - 2KCI(s) + 302(9) If 176.0 g of potassium chlorate reacts in a 2.90-L vessel, which was initially evacuated, what pressure of oxygen will be attained when the temperature is finally cooled to 25°C? Use the preceding chemical equation, and ignore the volume of solid product. Pressure = atm O2
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[References)
Liquid oxygen was first prepared by heating potassium chlorate, KCIO3, in a closed vessel to obtain oxygen at high
pressure. The oxygen was cooled until it liquefied.
2KCIO3 (s) → 2KCI(s) + 302 (g)
If 176.0 g of potassium chlorate reacts in a 2.90-L vessel, which was initially evacuated, what pressure of oxygen will be
attained when the temperature is finally cooled to 25°C? Use the preceding chemical equation, and ignore the volume of
solid product.
Pressure =
atm O2
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Expert Solution
Chemical equation
Given chemical equation :
2 KClO3 (s) ----> 2KCl(s) + 3O2 (g)
According to balanced chemical equation 2 moles of Potassium chlorate is decompose to from 3 moles of Oxygen.
Mole relation gives relation between number of moles and molar mass
1 mol = molar mass in grams
Atomic mass (g/mol)
K = 39
Cl = 35.5
O = 16
Molar mass of KClO3 = 1 × atomic mass of K + 1 × atomic mass of Cl + 3 × atomic mass of O
= 1 × 39 g/mol + 1 × 35.5 g/mol + 3 × 16 g/mol
= 122.5 g/mol
Apply mole relation on KClO3:
1 mol KClO3 = 122.5 g
Molar mass of O2 = 2 × atomic mass of O
= 2 × 16 g/mol
= 32 g/mol
Apply mole relation on O2:
1 mol O2 = 32 g
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