How many moles of KClO3 are needed to form 2.8 L of O2, measured at STP, according to the following reaction: 2KClO3 ® 2KCl + 3O2
1 mol = 22.4 Liters at STP
so 2.8 L/(22.4 L/mol ) = 1/8 mol of O2 is formed
You need 2Mol of K... for 3 mol of O2
so 2/3 * 1/8 = 1/12 mol of KClO3 needed in perfect world
9? How did you get 9 ?????
To determine the number of moles of KClO3 needed to form 2.8 L of O2, measured at STP, we can use the ideal gas law equation:
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)
At STP (Standard Temperature and Pressure), the conditions are:
- Pressure = 1 atm
- Temperature = 273.15 K
Given that the volume of O2 is 2.8 L, we can substitute these values into the equation and solve for the number of moles of O2:
(1 atm) * (2.8 L) = n * (0.0821 L·atm/mol·K) * (273.15 K)
2.8 = n * 22.4148
Now, we have the number of moles of O2. However, the balanced equation shows that the ratio of KClO3 to O2 is 2:3. Therefore, we need to find the number of moles of KClO3 corresponding to the number of moles of O2.
According to the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. So, we can set up a proportion:
2 moles KClO3 / 3 moles O2 = X moles KClO3 / 2.8 moles O2
Cross-multiplying the proportion, we get:
(2 moles KClO3) * (2.8 moles O2) = (3 moles O2) * (X moles KClO3)
Simplifying the equation:
5.6 = 3X
Dividing both sides by 3:
X = 5.6 / 3
X ≈ 1.87 moles
Therefore, approximately 1.87 moles of KClO3 are needed to form 2.8 L of O2, measured at STP.