Answer:
[tex]\large \boxed{0.102 \text{ g}}[/tex]
Explanation:
We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
Mᵣ: 74.55
2KClO₃ ⟶ 2KCl + 3O₂
V/mL: 50.0
1. Use the Ideal Gas Law to find the moles of O₂
[tex]\begin{array}{rcl}pV & = & nRT\\\text{1.00 atm} \times \text{0.0500 L} & = & n \times 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\0.0500 & = & 24.47n \text{ mol}^{-1}\\n & = & \dfrac{0.0500}{24.47\text{ mol}^{-1}}\\\\ & = & 2.044 \times 10^{-3} \text{ mol}\\\end{array}[/tex]
2. Calculate the moles of KCl
The molar ratio is 2 mol KCl:1 mol O₂
[tex]\rm \text{Moles of KCl} = 2.044 \times 10^{-3} \text{ mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = 1.363 \times 10^{-3} \text{ mol KCl}[/tex]
3. Calculate the mass of KCl
[tex]\text{Mass of KCl} = 1.363 \times 10^{-3} \text{ mol KCl} \times \dfrac{\text{74.55 g KCl}}{\text{1 mol KCl}} = \textbf{0.102 g KCl}\\\text{The mass of KCl produced is $\large \boxed{0.102 \text{ g}}$}[/tex]
