a) To find the normality of the hydrochloric acid, we need to know how many moles of HCl were used in the reaction. We can find this by using the balanced chemical equation for the reaction between HCl and Na2CO3:
2 HCl + Na2CO3 → 2 NaCl + CO2 + H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3. Therefore, the number of moles of HCl used in the reaction is:
moles HCl = (35.00 mL HCl) x (1 L / 1000 mL) x (N / 1 L)
where N is the normality of the HCl. We can solve for N:
N = (moles HCl / (35.00 mL)) x (1000 mL / 1 L) x (1 / 2)
N = (0.4800 g / 106 g/mol) x (1 mol Na2CO3 / 0.8) x (2 / 0.03500 L)
N = 0.685 N
Therefore, the normality of the hydrochloric acid is 0.685 N.
b) To find the molarity of the hydrochloric acid, we need to know the volume of the solution used in the reaction. We can assume that the density of the hydrochloric acid solution is 1.00 g/mL, which means that 35.00 mL of solution has a mass of 35.00 g. We can also calculate the number of moles of Na2CO3 used in the reaction:
moles Na2CO3 = 0.4800 g / (106 g/mol x 0.8)
moles Na2CO3 = 0.00567 mol
From the balanced equation, we know that 2 moles of HCl react with 1 mole of Na2CO3. Therefore, the number of moles of HCl used in the reaction is:
moles HCl = 2 x moles Na2CO3
moles HCl = 2 x 0.00567 mol
moles HCl = 0.0113 mol
Finally, we can calculate the molarity of the hydrochloric acid:
Molarity = moles HCl / volume of HCl solution (L)
Molarity = 0.0113 mol / (35.00 g / 1000 g/L) / (1.00 g/mL)
Molarity = 0.323 M
Therefore, the molarity of the hydrochloric acid is 0.323 M.