Question

how many grams of KCI(s)has been produced from the thermal decomposition of KCIO3(s) that produced 50.0ml of O2(g)at 25 degrees C and 1.00 atm pressure?

Answer 1

`\large \boxed{0.102 \text{ g}}`

Step-by-step 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

`\begin{array}{rcl}pV & = & nRT\\\text{1.00 atm} * \text{0.0500 L} & = & n * 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{298.15 K}\\0.0500 & = & 24.47n \text{ mol}^(-1)\\n & = & \frac{0.0500}{24.47\text{ mol}^(-1)}\\\\ & = & 2.044 * 10^(-3) \text{ mol}\\\end{array}`

2. Calculate the moles of KCl

The molar ratio is 2 mol KCl:1 mol O₂

`\rm \text{Moles of KCl} = 2.044 * 10^(-3) \text{ mol O}_(2) * \frac{\text{2 mol KCl}}{\text{3 mol O}_(2)} = 1.363 * 10^(-3) \text{ mol KCl}`

3. Calculate the mass of KCl

`\text{Mass of KCl} = 1.363 * 10^(-3) \text{ mol KCl} * \frac{\text{74.55 g KCl}}{\text{1 mol KCl}} = \textbf{0.102 g KCl}\\\text{The mass of KCl produced is $\large \boxed{0.102 \text{ g}}$}`