Can you show me how to do this?
2KClO3 -> 2KCl + 3O2
If 5.0(g) of KClO3 is decomposed , what volume of O2 is produced at STP?
5 g KClO3 = ?? mols.
??mols KClO3 will produce how many mols O2? Use the coefficients in the equation to do this conversion.
1 mol O2 occupies 22.4 L at STP. Convert mols you have to Liters.
Post any work if you get stuck and need further assistance. Please be specific about what you don't understand.
1.34
answer
To determine the volume of O2 produced at STP when 5.0 g of KClO3 is decomposed, let's follow these steps:
1. Convert the mass of KClO3 to moles.
To do this, you need to know the molar mass of KClO3, which can be calculated by adding up the atomic masses of potassium (K), chlorine (Cl), and oxygen (O) in the compound:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol (x3 since there are three oxygen atoms in KClO3)
Molar mass of KClO3 = (39.1 g/mol) + (35.5 g/mol) + (16.0 g/mol x 3) = 39.1 g/mol + 35.5 g/mol + 48.0 g/mol = 122.6 g/mol
Next, divide the mass of KClO3 given (5.0 g) by the molar mass of KClO3 to obtain the number of moles:
moles of KClO3 = 5.0 g / 122.6 g/mol = 0.041 mol
2. Determine the mole ratio between KClO3 and O2.
From the balanced chemical equation given:
2KClO3 -> 2KCl + 3O2
we can see that for every 2 moles of KClO3, 3 moles of O2 are produced.
3. Use the mole ratio to calculate the moles of O2 produced.
Multiply the moles of KClO3 (0.041 mol) by the mole ratio of O2/KClO3 (3/2) to find the moles of O2 produced:
moles of O2 = 0.041 mol x (3 mol O2 / 2 mol KClO3) = 0.062 mol
4. Convert moles of O2 to volume (in liters) at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.
Therefore, multiply the moles of O2 (0.062 mol) by the volume conversion factor to get the volume in liters:
volume of O2 = 0.062 mol x (22.4 L/1 mol) = 1.39 L
Therefore, when 5.0 g of KClO3 is decomposed, the volume of O2 produced at STP is approximately 1.39 L.
If you have any specific questions or get stuck at any step, let me know and I'll be happy to assist you further.