Calculation of NaCl Production from CaCO3 in a Chemical Reaction

To determine how many grams of sodium chloride (NaCl) are produced when 10.0 grams of calcium carbonate (CaCO3) are produced in a chemical reaction, we need to follow the principles of stoichiometry and understand the given reaction equation. This article will walk you through the process using two methods, making it easier to grasp the concept and solve similar problems.

Understanding the Reaction

The balanced chemical equation for the reaction between calcium carbonate and sodium carbonate is:

CaCl2(aq) Na2CO3(aq) → 2NaCl(aq) CaCO3(s)

This is a metathesis or partner exchange reaction. In this reaction, the insolvability of calcium carbonate (CaCO3) in aqueous solution drives the formation of a precipitate.

Stoichiometry and Molar Masses

Molar mass of CaCO3 40 12 3×16 100 g/mol Molar mass of NaCl 23 35.5 58.5 g/mol

From the balanced equation, we observe a 1:1 mole ratio between CaCO3 and NaCl. This means that one mole of CaCO3 reacts to form one mole of NaCl.

Method 1: Direct Calculation

Determine the number of moles of CaCO3 produced: Moles of CaCO3 10.0 g / 100 g/mol 0.1 mol Since the ratio of CaCO3 to NaCl is 1:1, the moles of NaCl formed are also 0.1 mol. Calculate the mass of NaCl formed: Mass of NaCl 0.1 mol × 58.5 g/mol 5.85 g

Therefore, 5.85 grams of NaCl are produced.

Method 2: Dimensional Analysis

The second method uses dimensional analysis to directly calculate the mass of NaCl:

10.0 g CaCO3 / 1 mol CaCO3 / 100 g CaCO3 × 1 mol NaCl / 1 mol CaCO3 × 58.5 g NaCl / 1 mol NaCl 10.0 / 100 × 58.5 5.85 g NaCl

This confirms that 5.85 grams of NaCl are produced.

Relevance to Given Information

It was previously mentioned that 10.0 grams of CaCO3 are produced. Based on the stoichiometry, the reaction would produce:

0.10 moles of CaCO3 0.20 moles of NaCl 12 grams of NaCl (given the molar mass of NaCl is 58.5 g/mol)

This confirms that the calculation aligns with the provided information and the balanced chemical equation.

Conclusion

Understanding stoichiometry is crucial in chemical reactions as it helps us predict the precise amounts of products formed from reactants. In this case, knowing the molar masses and the balanced equation allowed us to calculate the amount of NaCl produced from CaCO3. The methods used here can be applied to similar problems, ensuring accurate results and a deeper comprehension of chemical reactions.