Question

Complete and balance the equation for the thermal decompositon of potassium chlorate.

If 9.50 moles of oxygen is produced, how much heat is also produced? The heat of reaction is -89.4 kJ.

If you start with 307 grams of potassium chlorate, how many liters of oxygen will be produced at 723 torr and 32.0 °C?.

How much heat is produed when 307 grams of potassium chlorate is decomposed?

If the heat from the reaction was all absorbed by the 74.2 liter of collection water at 14.3 °C, what would the final temperature of the collection water?

Answer 1

1. The balanced equation for the thermal decomposition of potassium chlorate is:

2KClO3(s) → 2KCl(s) + 3O2(g)

2. Using the stoichiometry of the balanced equation, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, 9.50 moles of O2 are produced by (9.50 moles O2 / 3 moles O2 per 2 moles KClO3) = 6.33 moles KClO3. The heat produced by the decomposition of 6.33 moles of KClO3 is:

q = nΔHrxn = (6.33 mol)(-89.4 kJ/mol) = -566 kJ

3. To solve this problem, we need to use the ideal gas law to calculate the volume of O2 produced. First, we convert 307 g of KClO3 to moles:

n = m/M = 307 g / 122.55 g/mol = 2.50 mol KClO3

Using the stoichiometry of the balanced equation, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, 2.50 moles of KClO3 produce (3/2 x 2.50) = 3.75 moles of O2. Now we can use the ideal gas law to calculate the volume of O2 produced:

PV = nRT

V = nRT/P = (3.75 mol)(0.08206 L·atm/mol·K)(305.15 K)/(723 torr/760 torr/atm) = 8.59 L

4. The heat produced by the decomposition of 307 g of KClO3 is:

n = m/M = 307 g / 122.55 g/mol = 2.50 mol KClO3

q = nΔHrxn = (2.50 mol)(-89.4 kJ/mol) = -223 kJ

5. We can use the equation q = mcΔT to calculate the final temperature of the water. First, we need to calculate the heat capacity of the water:

C = mc = (74.2 L)(1.00 kg/L)(4.18 J/g·K) = 310 kJ/K

Now

Step-by-step explanation: