To solve this problem, we first need to use stoichiometry to determine the number of moles of O2 produced by the decomposition of 37.25 grams of KCIO3.
From the balanced equation, we know that 2 moles of KCIO3 decompose to produce 3 moles of O2. Therefore, we can use the molar mass of KCIO3 to convert 37.25 grams to moles:
37.25 g KCIO3 x (1 mol KCIO3/ 122.55 g KCIO3) = 0.3032 mol KCIO3
Using the stoichiometry of the balanced equation, we can convert the moles of KCIO3 to moles of O2:
0.3032 mol KCIO3 x (3 mol O2/ 2 mol KCIO3) = 0.4548 mol O2
Now that we know the number of moles of O2 produced, we can use the ideal gas law to calculate the volume of gas produced at the given temperature and pressure.
V = nRT/P
Where:
V = volume of gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature (in Kelvin)
P = pressure (in atm)
First, we need to convert the temperature to Kelvin:
12.43°C + 273.15 = 285.58 K
Now we can substitute the values into the ideal gas law equation:
V = (0.4548 mol)(0.0821 L·atm/K·mol)(285.58 K)/(941.34 kPa/101.325 kPa/atm)
V = 0.0151 L or 15.1 mL (rounded to three decimal places)
Therefore, 37.25 grams of KCIO3 will produce 15.1 mL of O2 at a temperature of 12.43°C and 941.34 kPa.