What volume of oxygen at s.t.p will be obtained when 24.5g of potassium trioxochlorate (v) crystals are heated?
how many moles in 24.5g?
per the reaction equation, how many moles of O2 are produced?
1 mole occupies 22.4L at STP
what is the volume of oxygen at s t p will be obtained when 24.5g of potassium trioxochlorite
To calculate the volume of oxygen gas obtained when potassium trioxochlorate (V) crystals are heated at standard temperature and pressure (STP), we can use the ideal gas law equation: PV = nRT. However, before we apply this equation, we need to determine the number of moles of potassium trioxochlorate (V) crystals and convert it into moles of oxygen gas.
Here are the steps to solve the problem:
1. Determine the formula of potassium trioxochlorate (V) crystals.
The formula of potassium trioxochlorate (V) is KClO₃.
2. Calculate the molar mass of KClO₃.
The molar mass of K is 39.10 g/mol, Cl is 35.45 g/mol, and O is 16.00 g/mol.
Therefore, the molar mass of KClO₃ is: (39.10 g/mol) + (35.45 g/mol) + (3 * 16.00 g/mol) = 122.55 g/mol.
3. Convert the mass of KClO₃ to moles.
Given mass of KClO₃ = 24.5 g.
Number of moles of KClO₃ = (Given mass of KClO₃) / (Molar mass of KClO₃)
= 24.5 g / 122.55 g/mol
= 0.1998 mol (approximately 0.20 mol)
4. Determine the molar ratio between KClO₃ and O₂.
From the balanced chemical equation: 2 KClO₃ → 2 KCl + 3 O₂
The stoichiometric ratio of KClO₃ to O₂ is 2:3.
5. Calculate the number of moles of O₂.
Number of moles of O₂ = (Number of moles of KClO₃) * (Molar ratio of O₂)
= 0.20 mol * (3/2)
= 0.30 mol
6. Apply the ideal gas law equation to calculate the volume of O₂ at STP.
STP conditions: P = 1 atm, T = 273 K, R = 0.0821 L·atm/(mol·K)
Rearrange the ideal gas law equation to solve for volume (V):
V = (n * R * T) / P
Substituting the values:
V = (0.30 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
≈ 6.6415 L
≈ 6.64 L (rounded to two decimal places)
Therefore, approximately 6.64 liters of oxygen gas would be obtained at STP when 24.5 g of potassium trioxochlorate (V) crystals are heated.