given the following equation 2ClO3---> 2 KCl + 3 O2
what volume does O2 occupied, at STP, if 34.0g of KClO3 reacts?
To find the volume of O2 gas produced, we can use the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given:
Mass of KClO3 = 34.0 g
1. First, we need to find the number of moles of KClO3 using its molar mass.
The molar mass of KClO3 is:
Molar mass of K = 39.10 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of KClO3 = (39.10 + 35.45 + (3 * 16.00)) g/mol
Molar mass of KClO3 = 122.55 g/mol
Number of moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
Number of moles of KClO3 = 34.0 g / 122.55 g/mol
2. Next, we need to determine the stoichiometry of the reaction.
From the equation, we can see that for every 2 moles of KClO3 that react, we produce 3 moles of O2.
3. Now, we can calculate the number of moles of O2 gas produced.
Number of moles of O2 = (Number of moles of KClO3) * (3 moles of O2 / 2 moles of KClO3)
4. Since the volume is asked at STP (Standard Temperature and Pressure), we can assume that the temperature (T) is 273.15 K and the pressure (P) is 1 atm.
5. Lastly, we can rearrange the Ideal Gas Law equation to solve for the volume (V):
V = (nRT) / P
Substituting the values into the equation will give us the volume of O2 gas produced.
To find the volume of O2 gas produced at STP (Standard Temperature and Pressure), we can use the ideal gas law. The ideal gas law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, we are given the mass of KClO3, so we need to convert it to moles. The molar mass of KClO3 can be calculated by adding up the atomic masses of potassium (K), chlorine (Cl), and oxygen (O). Potassium has an atomic mass of 39.10 g/mol, chlorine has an atomic mass of 35.45 g/mol, and oxygen has an atomic mass of 16.00 g/mol. Thus, the molar mass of KClO3 is:
(39.10 g/mol) + (35.45 g/mol) + (3 * 16.00 g/mol) = 122.55 g/mol
To convert grams of KClO3 to moles, we divide the mass by the molar mass:
34.0 g / 122.55 g/mol ≈ 0.2777 mol
From the balanced chemical equation, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, the number of moles of O2 produced can be calculated using the stoichiometry:
0.2777 mol KClO3 * (3 mol O2 / 2 mol KClO3) ≈ 0.4166 mol O2
Now, we have the number of moles of O2, and we can substitute it into the ideal gas law equation to find the volume. At STP, the temperature is 273.15 K and the pressure is 1 atm. The ideal gas constant, R, is 0.0821 L.atm/(mol.K). Therefore, the equation becomes:
V * (1 atm) = (0.4166 mol O2) * (0.0821 L.atm/(mol.K)) * (273.15 K)
Simplifying:
V = (0.4166 mol O2) * (0.0821 L.atm/(mol.K)) * (273.15 K) / (1 atm)
V ≈ 9.07 L
Therefore, the volume of O2 gas produced at STP, when 34.0 g of KClO3 reacts, is approximately 9.07 liters.