Let's break down the problem step by step: ### Given Data: - Total mass of mixture = **5.14 g** - Volume of flask = **1.50 L** - Initial CO₂ pressure = **750 torr** - Temperature = **30.0 °C** (= 303.15 K) - Final CO₂ pressure after reaction = **230 torr** ### Reactions: - **BaO(s) + CO₂(g) → BaCO₃(s)** - **CaO(s) + CO₂(g) → CaCO₃(s)** But only *BaO* and *CaO* react with CO₂ to form their respective carbonates, consuming CO₂ in the process. --- ### Step 1: Calculate moles of CO₂ before and after reaction using Ideal Gas Law (PV = nRT): **R = 0.0821 L·atm/(mol·K)** #### Initial moles of CO₂: - P₁ = 750 torr = 0.987 atm (1 atm = 760 torr) - V = 1.50 L - T = 303.15 K \[ n_1 = \frac{PV}{RT} = \frac{0.987 \times 1.50}{0.0821 \times 303.15} = \frac{1.4805}{24.888} = 0.0595 \text{ mol} \] #### Final moles of CO₂: - P₂ = 230 torr = 0.303 atm \[ n_2 = \frac{0.303 \times 1.50}{0.0821 \times 303.15} = \frac{0.4545}{24.888} = 0.0183 \text{ mol} \] #### Moles of CO₂ consumed: \[ \Delta n = n_1 - n_2 = 0.0595 - 0.0183 = 0.0412 \text{ mol} \] --- ### Step 2: Let x = mass of CaO, y = mass of BaO \[ x + y = 5.14 \text{ g} \] Both react 1:1 with CO₂: \[ \text{CaO} + \text{CO}_2 \rightarrow \text{CaCO}_3 \\ \text{BaO} + \text{CO}_2 \rightarrow \text{BaCO}_3 \] Let n(CaO) = a, n(BaO) = b \[ a + b = 0.0412 \text{ mol} \] - Molar mass CaO = 56.08 g/mol - Molar mass BaO = 153.33 g/mol \[ x = 56.08a \\ y = 153.33b \] \[ x + y = 5.14 \implies 56.08a + 153.33b = 5.14 \] And: \[ a + b = 0.0412 \] --- ### Step 3: Solve the System of Equations From \( a + b = 0.0412 \), so \( a = 0.0412 - b \): Substitute into mass equation: \[ 56.08(0.0412 - b) + 153.33b = 5.14 \\ 2.308 + 153.33b - 56.08b = 5.14 \\ 2.308 + 97.25b = 5.14 \\ 97.25b = 5.14 - 2.308 = 2.832 \\ b = \frac{2.832}{97.25} = 0.0291 \] So, \[ a = 0.0412 - 0.0291 = 0.0121 \] --- ### Step 4: Find Mass of Each \[ x = 56.08 \times 0.0121 = 0.679 \text{ g} \] \[ y = 153.33 \times 0.0291 = 4.464 \text{ g} \] Total: \( 0.679 + 4.464 = 5.143 \) g (matches total mass). --- ### Step 5: Percentages \[ \% \text{CaO} = \frac{0.679}{5.14} \times 100\% = 13.2\% \] \[ \% \text{BaO} = \frac{4.464}{5.14} \times 100\% = 86.8\% \] --- ### **Final Answer:** **Closest match:** **a) 13.4% CaO, 86.6% BaO** So the correct answer is **a**.