The reaction for the decomposition of potassium chlorate which is shown below. 2KClO3 (s) 3O2(g) + 2KCl(s) + 78 kJ What mass of KClO3 (s) would have decomposed to cause 250 mL of water in a calorimeter to change temperature from 20.00C to 55.00C. (5 marks)

To solve this problem, we need to use the concept of heat transfer and stoichiometry. Here are the steps:

1. First, we need to calculate the amount of heat absorbed by the water. We can use the formula q = mcΔT, where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity of water is 4.18 J/g°C. The mass of the water is its volume times its density, which is 1 g/mL for water. So, m = 250 mL * 1 g/mL = 250 g. The change in temperature, ΔT, is 55.00°C - 20.00°C = 35.00°C. So, q = (250 g)(4.18 J/g°C)(35.00°C) = 36675 J.

2. The heat released by the decomposition of KClO3 is given as 78 kJ per 2 moles of KClO3. We need to convert this to J per mole for easier calculation. So, the heat released per mole of KClO3 is (78 kJ * 1000 J/kJ) / 2 = 39000 J/mol.

3. Now, we can calculate the moles of KClO3 that would have decomposed to release the heat absorbed by the water. This is given by the heat absorbed by the water divided by the heat released per mole of KClO3. So, moles of KClO3 = 36675 J / 39000 J/mol = 0.94 mol.

4. Finally, we can calculate the mass of KClO3 that would have decomposed. This is given by the moles of KClO3 times its molar mass. The molar mass of KClO3 is 122.55 g/mol. So, mass of KClO3 = (0.94 mol)(122.55 g/mol) = 115.2 g.

So, approximately 115.2 g of KClO3 would have decomposed to cause the observed temperature change in the water.