How many grams of KClO3 must react to form 42.0 mL O2?
In order to determine the grams of KClO3 required to form 42.0 mL of O2, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction between KClO3 and O2 is:
2 KClO3 -> 2 KCl + 3 O2
From the equation, we can see that 2 moles of KClO3 react to form 3 moles of O2.
To solve the problem, we need to follow these steps:
Step 1: Convert milliliters of O2 to moles
Using the ideal gas law equation, we can convert the volume of O2 to moles:
PV = nRT
Where:
P = pressure (usually given in ATM)
V = volume (42.0 mL or 42.0 cm^3)
n = moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (usually given in Kelvin)
Since the pressure and temperature are not provided in the question, we need to assume standard conditions of 1 ATM and 298 K:
(1 ATM)(42.0 cm^3) = (n)(0.0821 L·atm/(mol·K))(298 K)
n = (1 ATM)(42.0 cm^3) / (0.0821 L·atm/(mol·K))(298 K)
n = 1.70421 x 10^-3 mol
Step 2: Use stoichiometry to determine the moles of KClO3
From the balanced equation, we know that 2 moles of KClO3 reacts with 3 moles of O2. Therefore, we can set up a ratio:
2 mol KClO3 / 3 mol O2 = x mol KClO3 / 1.70421 x 10^-3 mol O2
x = (2 mol KClO3 / 3 mol O2)(1.70421 x 10^-3 mol O2)
x = 1.13614 x 10^-3 mol KClO3
Step 3: Convert moles of KClO3 to grams
Finally, we can use the molar mass of KClO3 to convert moles to grams:
Molar mass KClO3 = (39.1 g/mol) + (35.5 g/mol) + (3(16.0 g/mol)) = 122.55 g/mol
Mass = (1.13614 x 10^-3 mol KClO3)(122.55 g/mol)
Mass = 0.1391 g
Therefore, approximately 0.1391 grams of KClO3 must react to form 42.0 mL of O2.