What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133°C and 0.880 atm? (The other reaction product is solid KCl.) (R = 0.08206 L·atm·K-1·mol-1)
To solve this problem, we first need to determine the moles of oxygen gas needed to produce 126 L at 133°C and 0.880 atm.
Using the ideal gas law: PV = nRT
We can rearrange the equation to solve for n (moles):
n = PV / RT
n = (0.880 atm) * (126 L) / (0.08206 L·atm·K-1·mol-1 * 406 K)
n ≈ 4.984 moles of O2
From the balanced chemical equation for the decomposition of KClO3:
2 KClO3 → 2 KCl + 3 O2
We know that for every 2 moles of KClO3 decomposed, we will produce 3 moles of O2. Therefore, the number of moles of KClO3 needed to produce 4.984 moles of O2 will be:
4.984 moles O2 * (2 moles KClO3 / 3 moles O2) ≈ 3.3227 moles KClO3
Now, we need to convert moles of KClO3 to grams using the molar mass of KClO3:
Molar mass of KClO3 = 39.09 g/mol (K) + 35.45 g/mol (Cl) + 3 * 16.00 g/mol (O) = 122.55 g/mol
Mass of KClO3 = 3.3227 moles KClO3 * 122.55 g/mol ≈ 407.73 grams
Therefore, about 407.73 grams of KClO3 must be decomposed to produce 126 L of oxygen gas at 133°C and 0.880 atm.