Oxygen gas can be generated by heating KClO3 in the presence of a catalyst.
KClO3 ----> KCl + O2 (unbalanced)
What volume (in L) of O2 gas will be generated at T = 25 oC and P = 10.26x104 Pa from 1.28 g KClO3?
1. Balance the equation.
2. Convert 1.28 g KClO3 to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles KClO3 to moles O2.
4. Convert moles O2 to L using PV = nRT. Don't forget to convert Pa to atm and T to Kelvin.
To determine the volume of oxygen gas generated from 1.28 g of KClO3, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (in Pa)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(K·mol) or 8.314 L·kPa/(K·mol))
T = temperature (in Kelvin)
First, let's calculate the number of moles of KClO3 using its molar mass:
Molar mass of KClO3 = 39.1 g/mol (K) + 35.45 g/mol (Cl) + 3 * 16 g/mol (O)
= 122.55 g/mol
Number of moles of KClO3 = mass of KClO3 / molar mass of KClO3
= 1.28 g / 122.55 g/mol
= 0.01045 moles
Since the balanced equation shows a 1:1 mole ratio between KClO3 and O2, we have 0.01045 moles of O2.
Now, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 °C + 273.15 = 298.15 K
Next, we can plug the values into the ideal gas law equation to find the volume of O2 gas:
PV = nRT
V = (nRT) / P
V = (0.01045 mol * 0.0821 L·atm/(K·mol) * 298.15 K) / (10.26x104 Pa)
Converting pressure from pascal (Pa) to kilopascals (kPa):
P = 10.26x104 Pa = (10.26x104 Pa) / (1000 Pa/kPa) = 10.26x10 kPa
V = (0.01045 mol * 0.0821 L·atm/(K·mol) * 298.15 K) / (10.26x10 kPa)
V = 0.0192 L
Therefore, the volume of oxygen gas generated from 1.28 g of KClO3 at T = 25 °C and P = 10.26x10^4 Pa is approximately 0.0192 L.