How many mL of O₂ gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.500 grams of KClO₃(s)according to the chemical equation shown below? 2 KClO₃(s)→ 2 KCl(s)+ 3 O₂(g) A) 50.2 mL B) 67.0 mL C) 151 mL D) 302 mL

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of KClO3 used. The molar mass of KClO3 is approximately 122.55 g/mol. Thus, 0.500 grams of KClO3 is equivalent to 0.500 g / 122.55 g/mol = 0.00408 mol. According to the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, 0.00408 mol of KClO3 will produce (3/2) * 0.00408 = 0.00612 mol of O2.Next, we use the ideal gas law equation PV = nRT to find the volume of O2 produced. We rearrange the equation to solve for V: V = nRT/P. We need to ensure that R is in the correct units, so we use R = 0.0821 L·atm/(mol·K), T is in Kelvin (25°C + 273 = 298 K), and P is in atm (755 mm Hg * (1 atm / 760 mm Hg) ≈ 0.994 atm).Substituting the values into the equation gives us V = (0.00612 mol) * (0.0821 L·atm/(mol·K)) * (298 K) / (0.994 atm) ≈ 0.151 L or 151 mL.

How Many ML of O2 Gas at 25°C and 755

How Many ML of O₂ Gas at 25°C and 755