Question

3g of impure Na2Co3 were made up to 250cm3 of solution .25cm3 required 21.0cm3 of 0.105M HCL for complete neutraiization a.write down the balanced chemical equation for the reaction b.calculate the molarity of pure Na2Co3 c.caiculate concentration of pure Na2Co

Answer 1

[AI-generated answer — Bot GPT 3.5]

a. Balanced chemical equation:

Na2CO3 + 2HCl -> 2NaCl + H2O + CO2

b. Calculation of the molarity of pure Na2CO3:
Moles of HCl used = 0.105 mol/L x 21.0/1000 L = 0.002205 mol
From the balanced chemical equation, 1 mole of Na2CO3 reacts with 2 moles of HCl.
Therefore, moles of Na2CO3 = 0.002205 mol / 2 = 0.0011025 mol
Volume of solution = 250 cm3 = 0.25 L
Molarity of pure Na2CO3 = 0.0011025 mol / 0.25 L = 0.00441 mol/L

c. Calculation of concentration of pure Na2CO3:
Molar mass of Na2CO3 = 105.99 g/mol
Mass of Na2CO3 in 3g of impure sample = 3g
Therefore, moles of impure Na2CO3 = 3g / 105.99 g/mol = 0.0283 mol
From the balanced chemical equation, 1 mole of Na2CO3 reacts with 2 moles of HCl.
Therefore, moles of pure Na2CO3 = 0.0283 mol / 2 = 0.01415 mol
Volume of solution = 250 cm3 = 0.25 L
Concentration of pure Na2CO3 = 0.01415 mol / 0.25 L = 0.0566 mol/L

Therefore, the molarity of pure Na2CO3 is 0.00441 mol/L and the concentration of pure Na2CO3 is 0.0566 mol/L.