What volume of oxygen measured at s.t.p will be produced on heating 24.5g of potassium trioxochlorate (v)?
ACCURATE ANSWER
2Kclo>>>>>2kcl +3o^2
2kclo³ produce 2kcl+30²
what is the volume of oxygen measured at s+p will bi produced on heating 24.59 of potassium trioxochlorote(v)
To find the volume of oxygen produced when heating a given mass of a compound, we need to follow these steps:
1. Determine the moles of potassium trioxochlorate (KClO3):
- The molar mass of KClO3 is calculated by adding the atomic masses of its constituent elements:
M(K) + M(Cl) + 3 * M(O) = 39.1 + 35.5 + 3 * 16 = 122.5 g/mol
- The number of moles of KClO3 can be calculated by dividing the given mass by its molar mass:
24.5 g / 122.5 g/mol = 0.2 mol
2. Write a balanced equation for the reaction:
KClO3 -> KCl + 3/2 O2
3. Determine the stoichiometry of the reaction:
From the balanced equation, we can see that 1 mole of KClO3 produces 3/2 moles of O2.
4. Calculate the moles of oxygen produced:
The moles of oxygen can be calculated by multiplying the moles of KClO3 by the stoichiometric ratio:
0.2 mol * (3/2) = 0.3 mol
5. Convert moles to volume at STP (standard temperature and pressure):
At STP, 1 mole of any gas occupies 22.4 liters of volume.
So, the volume of oxygen produced can be calculated by multiplying the number of moles by the molar volume:
0.3 mol * 22.4 L/mol = 6.72 L
Therefore, the volume of oxygen produced at STP when heating 24.5g of KClO3 is 6.72 liters.
2kclo³ produce 2kcl+3o²
2KClO3>>>2KCl + 3O2
moles of O2>3/2 moles of KClO3
= 3/2 * 24.5/123
volume= molegas*22.4