What volume of oxygen at STP is produced on heating 42.02gram of potassium trioxochlo rate( v)
To determine the volume of oxygen produced, we can use the ideal gas law. The formula for the ideal gas law is:
PV = nRT
Where:
P = pressure (at STP, P = 1 atm)
V = volume of gas
n = number of moles of gas
R = ideal gas constant (0.0821 L*atm/mol*K)
T = temperature (at STP, T = 273.15 K)
First, we need to calculate the number of moles of potassium trioxochlorate (KClO3). The molar mass of KClO3 is:
K = 39.10 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol
Molar mass of KClO3 = 39.10 + 35.45 + (3 * 16.00) = 122.55 g/mol
Number of moles of KClO3 = mass of KClO3 / molar mass of KClO3
= 42.02 g / 122.55 g/mol
≈ 0.3432 moles
According to the balanced equation of the reaction, KClO3 decomposes to form 3 moles of O2 gas for every mole of KClO3.
Therefore, the number of moles of oxygen produced = 3 * number of moles of KClO3
= 3 * 0.3432
= 1.0296 moles
Now, we can use the ideal gas law to calculate the volume of oxygen produced:
PV = nRT
(1 atm) * V = (1.0296 moles) * (0.0821 L*atm/mol*K) * (273.15 K)
V = (1.0296 * 0.0821 * 273.15) / 1
V ≈ 22.4 L
Therefore, approximately 22.4 liters of oxygen is produced at STP when heating 42.02 grams of potassium trioxochlorate.