Oxygen can be generated from the thermal decomposition of potassium chlorate, by the following reaction:2 KClO3(s) à 2 KCl(s) + 3 O2(g)What is the volume of oxygen produced, in litres (to 1 decimal place, input value only), at 173.9 kPa and 36.6 °C, when 27.4 grams of potassium chlorate are decomposed

To solve this problem, we need to use the ideal gas law and stoichiometry. Here are the steps:

1. First, we need to find the number of moles of KClO3 in 27.4 grams. The molar mass of KClO3 is approximately 122.55 g/mol. So, the number of moles = 27.4 g / 122.55 g/mol = 0.223 mol.

2. From the balanced chemical equation, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, 0.223 moles of KClO3 will produce 0.223 * (3/2) = 0.335 moles of O2.

3. Now we can use the ideal gas law to find the volume. The ideal gas law is 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 Kelvin. We need to convert the temperature from Celsius to Kelvin by adding 273.15, so T = 36.6 + 273.15 = 309.75 K.

4. The ideal gas constant R can be expressed in different units. Here we need to use the value that is compatible with the pressure in kPa and the volume in liters, which is R = 8.314 L·kPa/(K·mol).

5. Now we can solve the ideal gas law for V: V = nRT / P = 0.335 mol * 8.314 L·kPa/(K·mol) * 309.75 K / 173.9 kPa = 5.0 L.

So, the volume of oxygen produced is 5.0 liters.