Question

HCl reacts with Na2CO3 forming NaCl, H2O and CO2. This equation is balanced as written. 2HCl(aq) + Na2Co3 (aq) -> 2NaCl (aq) + H2O(l) + CO2(g).............................Part A---- what volume of 2.25 M HCl in liters is nedded to react completely (with nothing left over) with 0.500 L of 0.400 M Na2Co3. Use 3 sig figs and units. Part B---- A 317 mL sample of unknown HCl solution reacts completely with Na2CO3 to form 16.1 grams of CO2. What was the concentration of HCl solution. Use 3 sig figs and units

Answer 1

Reaction is

2HCl_(aq) + Na₂CO_3(aq) 2NaCl_(aq) + H₂O_(l) + CO_2(g)

a) From reaction

2 moles of HCl reacts with 1 mol of Na₂CO₃ to form product

Now Moles of Na₂CO₃ = 0.500 L x 0.400 M = 0.200 moles

So 0.200 moles of Na₂CO₃ will react with 0.400 moles of HCl

So required HCl = 0.400 moles / 2.25 M

= 0.178 L =178 mL

B)

Lets see moles of CO₂

Moles = mass /molar mass = 16.1 g / 44 g/mol

= 0.366 moles of CO₂

From reaction 2 moles of HCl reacts with 1 mole of Na₂CO₃ to form 1 mole of CO₂

therefore 0.366 x 2 = 0.732 moles of HCl is reacted

Now

Molarity = moles / Liter of solution

M = 0.732 moles / 0.317 L

= 2,31 M

So Concentration = 2.31 M