What volume of dry oxygen gass measured at s.t. p will produced from the decomposition of 40.5g of potassium trioxochlorate(v)?
(K=39,cl=35.5,o=16)
Bari
Solve
2KClO3 ==> 2KCl + 3O2
mols KClO3 = 40.5/122.5 = 0.3 estimated
0.3 mol KClO3 x (3 mols O2/2 mol KClO3) = 0.3 x 3/2 = 0.5 estimated
Then 0.5 mol O2 x 22.5 L/mol = approximate ?
You should go through and recalculate each since my numbers are just approximate. Post your work if you get stuck.
To determine the volume of dry oxygen gas produced from the decomposition of potassium trioxochlorate(V), you will need to use the stoichiometry of the balanced chemical equation. Here's how to calculate it step by step:
1. Start by writing the balanced equation for the decomposition of potassium trioxochlorate(V):
2KClO3(s) -> 2KCl(s) + 3O2(g)
This balanced equation shows that for every 2 moles of KClO3, 3 moles of O2 gas are produced.
2. Calculate the number of moles of KClO3 using its molar mass:
Molar mass of KClO3 = (2 * atomic mass of K) + atomic mass of Cl + (3 * atomic mass of O)
= (2 * 39) + 35.5 + (3 * 16)
= 78 + 35.5 + 48
= 161.5 g/mol
Number of moles of KClO3 = mass of KClO3 / molar mass
= 40.5 g / 161.5 g/mol
≈ 0.251 moles
3. Since the stoichiometry of the balanced equation is 2:3 (KClO3:O2), we can use this ratio to calculate the number of moles of O2 gas produced:
Number of moles of O2 = (3/2) * moles of KClO3
= (3/2) * 0.251 moles
≈ 0.3765 moles
4. Now, we need to convert the number of moles of O2 gas to its volume at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 liters.
Volume of O2 gas at STP = moles of O2 * molar volume of gas (at STP)
= 0.3765 moles * 22.4 L/mol
≈ 8.43 liters
Therefore, the volume of dry oxygen gas produced from the decomposition of 40.5g of potassium trioxochlorate(V) is approximately 8.43 liters at STP.