What volume of oxygen at STP can be produced by 6.125 g of KClO3(Potassium Chlorate) according to the equation, 2 KClO3 2KCl + 3O2 ( Atomic mass of K = 39)
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Answer:
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Answer:
Balanced equation
2KClO3 → 2KCl + 3O2
First calculate the moles of KClO3. Then use stoichiometry to calculate the moles of O2 that can be produced. Lastly, multiply the moles of O2 by the molar mass of a gas at STP, or use the ideal gas law.
Calculate moles KClO3.
n = m/M
n = mole = ? mol
m = mass (g) = 6.125 g KClO3
M = molar mass (g/mol) = (1 × 39.098 g/mol K) + (1 × 35.343 g/mol Cl) + (3 × 15.999 g/mol O) = 122.548 g KClO3/mol KClO3
n KClO3 = 6.125 g KClO3/122.548 g KClO3/mol KClO3 = 0.04998 mol KClO3
Calculate the moles of O2 that van be produced.
Multiply mol KClO3 by the mole ratio between KClO3 and O2 in the balanced equation so that mol KClO3 cancels, leaving mol O2.
0.04998 mol KClO3 × 3 mol O2/2 mol KClO3 = 0.07497 mol O2
Calculate the volume of O2 at STP.
Multiply mol O2 by the molar volume of a gas at STP.
At STP defined as 273.15 K (0°C) and 1 atm, the molar volume of a gas is 22.414 L/mol.
0.07497 mol O2 × 22.414 L/mol = 1.680 L O2 to four significant figures
At STP defined as 273.15 K (0°C) and 10^5 Pa (100 kPa), the molar volume of a gas is 22.711 L/mol.
0.07497 mol O2 × 22.711 L/mol = 1.703 L to four significant figures
You can also use the ideal gas law, but you will get the same answers.
PV = nRT
Solve for volume.
R = gas constant
R = 0.082057 L·atm/K·mol or
R = 8.314 L·kPa/K·mol