Question

a 655- ml sample of unknown hcl solution reacts completely with na2co3 to form 14.1 g co2 . what was the concentration of the hcl solution?

Answer 1

The concentration of the HCl solution is found to be 0.978 M after calculating the number of moles of CO2 produced and using the stoichiometry of the reaction between HCl and Na2CO3.

Step-by-step explanation:

To determine the concentration of an HCl solution that reacted with Na2CO3 to produce 14.1 g of CO2, first, we calculate the number of moles of CO2 produced using its molar mass (44.01 g/mol):

n(CO2) = mass/molar mass = 14.1 g / 44.01 g/mol = 0.3204 mol

The balanced chemical equation for the reaction between HCl and Na2CO3 is:

HCl + Na2CO3 → NaCl + CO2 + H2O

From the stoichiometry of the reaction, 1 mole of Na2CO3 reacts with 2 moles of HCl, so:

mols of HCl = 2 × mols of CO2 = 2 × 0.3204 mol = 0.6408 mol

To find the concentration of HCl, we divide the moles of HCl by the volume of the solution (in liters):

Concentration (M) = moles of solute / volume of solution (L)

Convert 655 mL to liters: 655 mL × 1 L/1000 mL = 0.655 L

Concentration of HCl = 0.6408 mol / 0.655 L = 0.978 M