Potassium chlorate decomposes to potassium chloride and oxygen. If 20.8 g of potassium chlorate decomposes, how many liters of oxygen will form at STP?

5.71 L of oxygen will be formed at STP

Explanation:

Step 1: Data given

Mass of potassium chlorate = 20.8 grams

Molar mass potassium chlorate = 122.55 g/mol

Step 2: The balanced equation

2KClO3 → 2KCl + 3O2

Step 3: Calculate moles KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3 = 20.8 grams / 122.55 g/mol

Moles KClO3 = 0.170 moles

Step 4: Calculate moles O2

For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

For 0.170 moles KClO3 we'll have 3/2 *0.170 = 0.255 moles O2

Step 5: Calculate volume O2

1 mol = 22.4 L

0.255 moles = 5.71 L

5.71 L of oxygen will be formed at STP

relative humidity is a measure of the amount of moisture in the air relative to the total amount of moisture the air can hold.

so d

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :