Question

Problem 2. Consider the following reaction, which proceeds to completion 1 Tico3(aq) + 2 HCIO3(aq) → 1 Ti(C103)2(aq) + 1 CO2(g) + 1 H2O(1) 24.73 g of Tico, reacts with 623 mL of 0.756 M HCİOg The temperature is 15.5 °C and total pressure is 18.1 psi? The vapor pressure of water at this temperature is 13.31 torr. a. Write the net ionic equation. b. Which ion is limiting (use the net ionic equation)? c. How many moles of the excess ion remain? d. How many moles of each spectator ion remain? e. What is the partial pressure of CO2, in atm? f. What volume of dry CO2, in cm3, is produced?

please answer all parts of the problem

Answer 1

(a) Total ionic reaction is:

Ti²⁺(aq) + CO₃^2-(aq) + 2H⁺(aq) + 2ClO₃^-(aq)---> Ti²⁺(aq) + 2ClO₃^-(aq) +CO₂(g) + H₂O(l)

Net Ionic reaction :

CO₃^2-(aq) + 2H⁺(aq) --->CO₂(g) + H₂O(l)

(b) mass of TiCO₃ = 24.73 g

Now, 24.73 g TiCO₃ * ( 1mol / 107.88 g) * ( 2mol HClO₃/ 1 mol TiCO₃) = 0.458 mol

Moles of HClO₃ given = Molarity * volume = 0.756 M *(623/1000)L = 0.471 mol

So, TiCO₃ is a limiting reagent

Hence, CO₃^2- is a limiting reagent

(c) Mole of HClO₃ remains = 0.471 - 0.458 = 0.0130 moles

Thus, we have 0.013* moles of H⁺ are in excess.

(d) Spectator ions are those which remains in solution and are in same form on both side of reaction

0.458 mol of Ti²⁺ and 0.471 mol of ClO₃ ions remained.

(e) Here, only CO₂ is in gaseous form. So, pressure is only due to CO₂

Partial pressure of CO₂ = Total pressure - Pressure of water

Total pressure = 18.1 psi = 18.1psi * (1 atm / 14.696 psi) = 1.23 atm

Pressure of water = 13.31 torr = 13.31 torr * (1 atm / 760 torr) = 0.0175 atm

So, Partial pressure of CO₂ = 1.23 - 0.0175 = 1.21 atm

(f) Temperature (T) = 15.5^oC = 15.5 +273.15 = 288.65 K

Pressure (P) = 1.21 atm

R = gas constant = 0.0821 L.atm / mol.K

Moles of CO₂ formed = Moles of TiCO₃ reacted = 0.458 mol

From Ideal gas equation

PV = nRT

V = nRT/P = (0.458 mol*0.0821 L.atm / mol.K * 288.65 K) / 1.21 atm = 8.95 L

1L = 1000 cm³

So, Volume of CO₂ produced = 8950 cm³