Calculate the volume of oxygen that would be produced if 12.35g of potassium trioxocarbonate iv is decmposed using:
2Kclo3➡2kcl + 3o2
To calculate the volume of oxygen produced, we need to use the balanced chemical equation and the ideal gas law.
First, we need to convert the given mass of potassium trioxocarbonate IV (KClO3) to moles. The molar mass of KClO3 is 122.55 g/mol.
Number of moles of KClO3 = 12.35 g / 122.55 g/mol
Number of moles of KClO3 = 0.1007 mol
According to the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, the number of moles of O2 produced can be calculated using the following ratio:
Number of moles of O2 = (0.1007 mol KClO3) * (3 mol O2 / 2 mol KClO3)
Number of moles of O2 = 0.1511 mol
Now, we can use the ideal gas law to calculate the volume of the oxygen gas. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (in Kelvin)
Assuming the conditions are at STP (Standard Temperature and Pressure), where T = 273 K and P = 1 atm, we can substitute these values into the ideal gas law equation. Therefore:
V = (0.1511 mol * 0.0821 L•atm/mol•K * 273 K) / 1 atm
V = 3.755 L
Therefore, the volume of oxygen gas produced is approximately 3.755 liters.