When the following reaction took place at 23.8 and a pressure of 758 torr, 3.9 L of o2 were collected. What mass of KCIO3 decomposed?
KCIO3->KCI+O2
Use PV = nRT and solve for n = mols O2 at the conditions listed. Use that and simple stoichiometry to solve for grams KClO3.
To determine the mass of KCIO3 decomposed, we need to use the ideal gas law equation and stoichiometry.
First, let's consider the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = gas constant (0.0821 L*atm/mol*K)
T = temperature (in Kelvin)
Now, let's convert the given pressure to atm and the given volume to moles:
Given:
Pressure = 758 torr = 758/760 atm ≈ 0.9955 atm
Volume of O2 = 3.9 L
Using the ideal gas law equation, we can calculate the moles of O2:
n = PV / RT
n = (0.9955 atm * 3.9 L) / (0.0821 L*atm/mol*K * 298 K)
n ≈ 0.1625 moles of O2
Now, let's use the stoichiometry to determine the moles of KCIO3 decomposed.
From the balanced chemical equation: 1 mole of KCIO3 produces 1 mole of O2.
Therefore, the moles of KCIO3 decomposed is also 0.1625 moles.
Finally, to calculate the mass of KCIO3 decomposed, we need to use the molar mass of KCIO3.
The molar mass of KCIO3 = (1 * atomic mass of K) + (1 * atomic mass of Cl) + (3 * atomic mass of O) ≈ 122.55 g/mol
Mass of KCIO3 decomposed = moles of KCIO3 decomposed * molar mass of KCIO3
Mass of KCIO3 decomposed ≈ 0.1625 moles * 122.55 g/mol ≈ 19.90 g
Therefore, the mass of KCIO3 decomposed is approximately 19.90 grams.
To determine the mass of KCIO3 that decomposed, we need to use the ideal gas law and stoichiometry.
First, let's convert the pressure from torr to atm. Since 1 atm = 760 torr, we divide the pressure by 760 torr to get:
758 torr ÷ 760 torr/atm = 0.997 atm
Next, we can use the ideal gas law equation, PV = nRT, to find the number of moles of O2 (n):
P = pressure in atm (0.997 atm)
V = volume in liters (3.9 L)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (23.8°C = 23.8 + 273.15 = 296.95 K)
n = (P × V) ÷ (R × T)
n = (0.997 atm × 3.9 L) ÷ (0.0821 L·atm/mol·K × 296.95 K)
n ≈ 0.157 mol
From the balanced chemical equation, we know that the molar ratio between KCIO3 and O2 is 1:1. This means that 1 mole of KCIO3 decomposes to produce 1 mole of O2.
Therefore, the mass of KCIO3 that decomposed is equal to the molar mass of KCIO3 multiplied by the number of moles of KCIO3:
Now, let's calculate the molar mass of KCIO3:
K (39.10 g/mol) + Cl (35.45 g/mol) + 3 × O (16.00 g/mol)
= 39.10 g/mol + 35.45 g/mol + 3 × 16.00 g/mol
≈ 122.55 g/mol
Mass of KCIO3 decomposed = molar mass of KCIO3 × number of moles of KCIO3
Mass of KCIO3 decomposed = 122.55 g/mol × 0.157 mol
Mass of KCIO3 decomposed ≈ 19.23 g
Therefore, approximately 19.23 grams of KCIO3 decomposed in the given reaction.