Calcium chloride and Sodium carbonate Reaction | CaCl₂ + Na₂CO₃

Calcium chloride (CaCl₂) reacts with sodium carbonate (Na₂CO₃) and form calcium carbonate (CaCO₃) and sodium chloride (NaCl). Calcium carbonate is a white precipitate and insoluble in water.

In this tutorial, we will discuss followings.

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Reaction of CaCl₂ and Na₂CO₃ and balanced equation

As mentioned earlier, calcium carbonate and sodium chloride are given as results. To give these products, an aqueous phase is required because cations and anions should be dissociated in water. In solid phase, free cations and anions are not available.

Both CaCl₂ and Na₂CO₃ are soluble in water and dissociates completely to ions. In aqueous solution, CaCl₂ dissociates to Ca²⁺ and Cl^- ions. As well, Na₂CO₃ dissociates to Na⁺ and CO₃^2- ions.

Balanced equation of CaCl₂ and Na₂CO₃

CaCl_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaCl_(aq)

When reaction performs, all reactants and products are in aqueous state. According to the balanced equation, one mole of CaCl₂ reacts with one mole of Na₂CO₃ and gives one mole of CaCO₃ and one mole of NaCl respectively.

Reactants of the reaction: CaCl₂ and Na₂CO₃

Calcium chloride (CaCl₂)

Calcium chloride is a white solid at room temperature and soluble in water to give a colourless aqueous solution. When CaCl₂ is dissolved in water, it dissociates to Ca²⁺ and Cl^- ions.

Sodium carbonate (Na₂CO₃)

Sodium carbonate is a white solid and soluble in water. Aqueous sodium carbonate solution is colourless and dissociates to Na⁺ and CO₃^2- ions.

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Products of the reaction: CaCO₃ and NaCl

Calcium carbonate (CaCO₃)

Calcium carbonate is insoluble in water and deposited as a white precipitate.

Sodium chloride (NaCl)

Sodium chloride is a white solid at room temperature and highly soluble in water.

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Change of oxidation numbers

Oxidation numbers of atoms are not changed during the reaction. Therefore, this reaction is not a redox reaction.

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Physical and chemical observation of CaCl₂ and Na₂CO₃ reaction

Here, we will see some physical observations and chemical properties changes during the reaction.

Colour and physical state changes

• Both CaCl₂ and Na₂CO₃ are colourless solutions.
• One product, CaCO₃ is a white precipitate while other product NaCl exist as in the aqueous phase. So, you can see a white solid at the bottom of the solution at the end of the reaction.

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Safety, health hazards and environmental impacts due to NaNO₂, HCl, HNO₂

• CaCl₂: Irritant
• Na₂CO₃: Irritant
• CaCO₃: Corrosive and irritant
• NaCl: An eye irritant, if large amounts are ingested toxic characteristics are possible

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Questions

Ask your chemistry questions and find the answers

You are given 0.01 mol dm^-3 CaCl₂ 50 cm³ aqueous solution and 0.01 mol Na₂CO₃ 50 cm³ aqueous solution. Both of these two solutions are mixed.

1. Write the balanced equation between CaCl₂ and Na₂CO₃
2. Calculate how much CaCO₃ is deposited in the aqueous solution.
3. Mention what assumptions are made by you during the calculations.

1. Balanced equation of CaCl₂ and Na₂CO₃

CaCl_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaCl_(aq)

2. Calculate how much CaCO₃ is deposited in the aqueous solution.

To decide how much CaCO₃ is formed, you should calculate followings.

• How much CaCl₂ and Na₂CO₃ are there in solutions?: Use n = CV equation.
• How much CaCl₂ and Na₂CO₃ are reacted: Apply Stoichiometric balanced equation.

Amount of CaCl₂

• Amount of CaCl₂ in CaCl₂ aqueous solution = 0.01 mol dm^-3 * 50 * 10^-3 dm³
• Amount of CaCl₂ in CaCl₂ aqueous solution = 0.0005 mol

Amount of Na₂CO₃

• Amount of Na₂CO₃ in Na₂CO₃ aqueous solution = 0.01 mol dm^-3 * 50 * 10^-3 dm³
• Amount of Na₂CO₃ in Na₂CO₃ aqueous solution = 0.0005 mol

Tabulated calculated values as below. According to the stoichiometric balanced equation, we can decide quantities of reacted (reactants) and quantities of generated (products).

So, all CaCl₂ and Na₂CO₃ are consumed during the reaction. We can calculate how much CaCO₃ is precipitated in the solution.

Amount of CaCO₃

• Amount of CaCO₃ = 0.05 mol * 100 g mol^-1
• Amount of CaCO₃ aqueous solution = 5 g

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