Final answer:
To determine how many grams of potassium chlorate were reacted to produce 5.30 liters of oxygen, we use the ideal gas law to find the moles of oxygen and then the stoichiometric relationship from the balanced chemical equation to find the moles of potassium chlorate. We then convert this to grams using the chemical's molar mass.
Step-by-step explanation:
The student has provided the balanced chemical equation for the decomposition of potassium chlorate (KClO3) to potassium chloride (KCl) and oxygen gas (O2): 2 KClO3 → 2 KCl + 3 O2. To find out how many grams of potassium chlorate were reacted to produce 5.30 liters of oxygen at 117°C and 0.995 atm, we need to use the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvins.
First, we convert the volume of oxygen to moles using the ideal gas law.
Given:
Volume (V) = 5.30 L
Temperature (T) = 117°C = 390.15 K (add 273.15 to convert to Kelvin)
P is the pressure = 0.995 atm.
R (ideal gas constant) = 0.0821 L·atm/K·mol
We then solve for n (number of moles of oxygen):
n = PV / RT = (0.995 atm × 5.30 L) / (0.0821 L·atm/K·mol × 390.15 K)
Once we have the moles of O2, we can use the stoichiometry from the balanced equation to find the moles of KClO3 needed. Since 3 moles of O2 are produced from 2 moles of KClO3, we calculate the moles of KClO3 required and then convert this to grams using the molar mass of KClO3 (122.55 g/mol as per the provided reference).
Finally, the calculated mass of KClO3 represents the theoretical yield of potassium chlorate that reacted to produce the given volume of oxygen gas under the specified conditions.