Chemistry Question: Neutralization Reaction Calculation

What is the minimum mass of anhydrous sodium carbonate required to neutralize 50.0 cm³ of 2.00 moldm³ HCl(aq)? To neutralize 50.0 cm³ of 2.00 M HCl(aq), you would need a minimum mass of 10.6 g of anhydrous sodium carbonate (Na2CO3).

Neutralization Reaction Calculation

Neutralization reactions in chemistry involve the reaction between an acid and a base to form salt and water. In this particular case, we are looking at the neutralization reaction between hydrochloric acid (HCl) and anhydrous sodium carbonate (Na2CO3).

The balanced chemical equation for the reaction is:

HCl(aq) + Na2CO3(s) → NaCl(aq) + CO2(g) + H2O(l)

The first step in determining the minimum mass of anhydrous sodium carbonate required is to calculate the number of moles of HCl present in the solution. This can be done using the formula:

Moles of HCl = Volume (in dm³) × Concentration (in mol/dm³)

Given that the volume of the HCl solution is 50.0 cm³ (which is equivalent to 0.0500 dm³) and the concentration is 2.00 mol/dm³, we can calculate the moles of HCl:

Moles of HCl = 0.0500 dm³ × 2.00 mol/dm³ = 0.100 mol

Since the molar ratio between HCl and Na2CO3 is 2:1, we can use this ratio to determine the minimum mass of Na2CO3 needed to neutralize the acid. The formula to calculate the mass of Na2CO3 is:

Mass of Na2CO3 = Moles of HCl × Molar mass of Na2CO3

By substituting the values, we get:

Mass of Na2CO3 = 0.100 mol × 105.99 g/mol = 10.6 g

Therefore, to neutralize 50.0 cm³ of 2.00 M HCl(aq), a minimum mass of 10.6 g of anhydrous sodium carbonate (Na2CO3) is required.

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