calculate the volume of oxygen at 15°c and 745mmHg pressure which could be obtain by heating 10g of potassium trioxochloride.
(molar volume at stp is 22.4dm3. K = 39, Cl = 35.5 O = 10)
Why do you think the right method of approach should be the ideal gas equation? Don't you think the right method of approach should be a mix of mass-volume relationship with general gas equation?
2KClO3 ==> 2KCl + 3O2
mols KClO3 = grams/molar mass = ?
mols O2 generated = mols KClO3 x 3/2 = ?
Use PV = nRT to solve for V in liters.
Post your work if you get stuck
To calculate the volume of oxygen, we need to determine the number of moles of oxygen gas produced from the given mass of potassium trioxochloride (KClO3). Let's break down the steps to solve this problem:
1. Determine the molar mass of KClO3.
- Molar mass of K = 39 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of KClO3 = (39 + 35.5 + (3 * 16)) g/mol = 122.5 g/mol
2. Convert the mass of KClO3 into moles.
- Moles of KClO3 = mass / molar mass = 10 g / 122.5 g/mol ≈ 0.082 mol
3. Use the balanced chemical equation to determine the mole ratio of oxygen to KClO3.
- The balanced equation for the decomposition of KClO3 is: 2 KClO3 --> 2 KCl + 3 O2
- From the equation, we see that 2 moles of KClO3 produce 3 moles of O2.
- So, the moles of O2 = 0.082 mol * (3 mol O2 / 2 mol KClO3) ≈ 0.123 mol
4. Apply the ideal gas law to find the volume of oxygen gas at the given temperature and pressure.
- The ideal gas law equation is: PV = nRT
- Rearrange the equation to solve for volume: V = (nRT) / P
- Given that n = 0.123 mol, P = 745 mmHg, T = 15°C + 273.15 (convert to Kelvin), and R = 0.0821 L·atm/(K·mol)
- V = (0.123 mol * 0.0821 L·atm/(K·mol) * (15°C + 273.15) K) / 745 mmHg
5. Convert the volume to dm³ (cubic decimeters):
- 1 L = 1 dm³, so V (in dm³) = V (in L)
Now, let's calculate the volume:
V = (0.123 mol * 0.0821 L·atm/(K·mol) * (15°C + 273.15) K) / 745 mmHg
V ≈ (0.123 * 0.0821 * 288.15) / 745 dm³
V ≈ 0.003981 dm³
Therefore, the volume of oxygen gas obtained from heating 10g of KClO3 at 15°C and 745 mmHg pressure is approximately 0.003981 dm³.