At STP potassium chlorate (KClO3) decomposes to produce solid potassium chloride (KCL) and oxygen gas (O2) according to the balanced chemical equation:
2KClO3(s)->2KCl(s)+3O2(g)
What volume of oxygen gas, measured at 40 degrees celsius and 85.4 kPa, will be produced when 13.5 g of potassium chlorate is decomposed? Answer is 5 L. what gas law should i use?
PV = nRT
From equation 1 mole KClO3 => 3/2 mole O2 at STP x 22.4L/mole = 33.6L at STP.
Given 13.5g KClO3/(122g/mol)=> 3.73L O2 at STP. Adjusting for 40-deg C and 85.4Kpa (~641mm) ...
Vol@40C&641mm
= 3.73(760/641)(313/273)L
= 5.07 Liters
If you can calculate moles O2 ~0.167mole O2 from 13.5g KClO2 and use 313K & (641/760)atm then, PV=nRT is a shorter route.
To solve this problem, you can use the combined gas law to calculate the volume of oxygen gas produced. The combined gas law states:
(P₁V₁)/T₁ = (P₂V₂)/T₂
where P₁, V₁, and T₁ represent the initial conditions (pressure, volume, and temperature), and P₂, V₂, and T₂ represent the final conditions.
In this case, the initial conditions are not given, so we can assume standard temperature and pressure (STP). At STP, temperature is 0 degrees Celsius or 273.15 Kelvin, and pressure is 101.3 kPa.
Given:
- The final conditions: temperature (T₂) = 40 degrees Celsius = 273.15 + 40 = 313.15 Kelvin and pressure (P₂) = 85.4 kPa
- The balanced chemical equation: 2KClO₃(s) -> 2KCl(s) + 3O₂(g)
- The molar mass of KClO₃ is 122.55 g/mol
To calculate the volume of oxygen gas produced, follow these steps:
Step 1: Convert the mass of KClO₃ to moles.
moles of KClO₃ = mass / molar mass of KClO₃
moles of KClO₃ = 13.5 g / 122.55 g/mol
Step 2: Calculate the number of moles of O₂ produced.
According to the balanced chemical equation, the molar ratio of KClO₃ to O₂ is 2:3. Therefore, the moles of O₂ produced will be 3/2 times the moles of KClO₃.
moles of O₂ = (3/2) * moles of KClO₃
Step 3: Use the ideal gas law to calculate the volume of O₂ at STP.
PV = nRT
V = (nRT) / P
V = [(3/2) * moles of KClO₃ * R * T] / P
Note: R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (273.15 K)
Substituting the given values:
V = [(3/2) * (13.5 g / 122.55 g/mol) * (0.0821 L·atm/mol·K) * 313.15 K] / 101.3 kPa
After solving the equation, the result will be the volume of oxygen gas produced in liters.
The final answer you provided (5 L) may be an approximation. Please double-check the calculations to ensure accuracy.