A567-ml sample of unknown hcl solution reacts completely with na2co3 to form 15.1 g co2. what was the concentration of the hcl solution?

1.21 m The balanced reaction is 2 HCl + Na2CO3 ==> CO2 + H2O + 2 NaCl which indicates that for every mole of CO2 produced, 2 moles of HCl are required. So let's determine how many moles of CO2 were produced, starting with the atomic weights of carbon and oxygen. Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Moles CO2 = 15.1 g / 44.0087 g/mol = 0.343113975 moles Because of the balanced equation, we know we needed twice as many moles of HCl, so 0.343113975 mol * 2 = 0.68622795 mol And since molarity is defined as moles per liter, just divide the number of moles by the number of liters, so: 0.68622795 mol / 0.567 l = 1.210278572 mol/l Since we only have 3 significant figures in our data, round to 3 figures, giving 1.21 m

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