The thermal decomposition of potassium chlorate can be used to produce oxygen in the laboratory.

What volume (L) of O2 gas at 25 °C and 1.00 atm pressure is produced by the decomposition of 7.5 g of KClO3 (s)?

Here is my work:

V=nRT/P

.0611995 mol of O2 x .0821Latm/molL x298.15K / 1.0atm=1.498

Why is 1.498 incorrect?

.0611995 mol of KClO3 needed to be converted to .09179925 mol of O2 b/c of the coeficients

The answer is correct except for not being labeled and not rounded off correctly.

The thermal decomposition of potassium chlorate can be used to produce oxygen in the laboratory.

2KClO3 (s) ® 2KCl (s) + 3O2 (g)

What volume (L) of O2 gas at 25°C and 1.00 atm pressure is produced by the decomposition of 7.5 g of KClO3 (s)?
Group of answer choices
4.5
3.7
7.5
11
2.2
THE ANSWER IS 2.24 ROUNDED TO 2.2

Your calculation is almost correct, but there seems to be a miscalculation in the number of moles of oxygen gas produced. Let's go through the correct steps to find the accurate answer:

1. Start with the given mass of potassium chlorate (KClO3): 7.5 g.

2. Calculate the number of moles of KClO3 using its molar mass. The molar mass of KClO3 is the sum of the atomic masses of its constituent elements (K: 39.10 g/mol, Cl: 35.45 g/mol, and O: 16.00 g/mol). Summing these values gives us:

Molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + 16.00 g/mol = 90.55 g/mol.

Moles of KClO3 = 7.5 g / 90.55 g/mol ≈ 0.0828 mol (rounded to four decimal places).

3. According to the balanced equation for the thermal decomposition of KClO3:

2 KClO3(s) → 2 KCl(s) + 3 O2(g).

This means that for every 2 moles of KClO3, 3 moles of O2 gas are produced.

4. Using this stoichiometry, convert the moles of KClO3 to moles of O2:

Moles of O2 = 0.0828 mol KClO3 × (3 mol O2 / 2 mol KClO3) ≈ 0.1242 mol (rounded to four decimal places).

5. Finally, use the ideal gas law equation: PV = nRT, where P is the pressure (1.00 atm), V is the volume (which we need to find), n is the number of moles of gas (0.1242 mol), R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin (25 °C = 298.15 K).

Rearranging the equation to solve for V, we have:

V = (nRT) / P = (0.1242 mol × 0.0821 L·atm/(mol·K) × 298.15 K) / 1.00 atm ≈ 2.915 L (rounded to three decimal places).

Based on the correct calculations, the volume of oxygen gas produced is approximately 2.915 L, not 1.498 L.