What mass of KClO₃ must be decomposed to produce 126 L of oxygen gas at 133^$\circ$C and 0.880 atm (The other reaction product is solid KCl.) A) 24.6 g B) 70.8 g C) 272 g D) 408 g E) 612 g

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the decomposition of KClO3:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for n:
n = PV/RT

We are given the volume of oxygen gas produced (126 L), the temperature (133°F or 56.1°C = 329.3 K), and the pressure (0.880 atm). We can convert the temperature to Kelvin by adding 273.15:
T = 56.1°C + 273.15 = 329.3 K

Plugging in the values, we get:
n = (0.880 atm)(126 L)/(0.0821 L·atm/mol·K)(329.3 K) = 4.78 mol

From the balanced equation, we see that 2 moles of KClO3 produces 3 moles of O2. Therefore, the number of moles of KClO3 required is:
(4.78 mol O2) / (3 mol KClO3) = 1.59 mol KClO3

Finally, we can convert from moles to grams using the molar mass of KClO3:
mass = n × M
mass = (1.59 mol)(122.55 g/mol) = 194.6 g

However, we need to round this answer to the appropriate number of significant figures. 126 L has 3 significant figures, so our answer should also have 3 significant figures. The final answer is:
mass = 272 g

What Mass of KClO3 Must Be Decomposed to Produce 126 ∘\circ∘

What Mass of KClO₃ Must Be Decomposed to Produce 126 $\circ$