Question

a 657-ml sample of unknown HCL solution reacts completely with Na2CO3 to form 11.1 g CO2. What was the concentration of the HCI solution?

Answer 1

The concentration of the HCI solution : 0.767 M

Further explanation

Reaction

Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)

mass of CO₂ = 11.1 g

mol of CO₂ (MW= 44,01 g/mol) :

`\tt mol=(mass)/(MW)\\\\mol=(11.1)/( 44,01)\\\\mol=0.252`

From the equation above, mol ratio of HCl : CO₂ = 2 : 1, so mol HCl :

`\tt mol~HCl=(2)/(1)* 0.252=0.504`

Molarity shows the number of moles of solute in every 1 liter of solution.

`\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}`

The molarity of unknown HCl :

mol=n=0.504

volume=V=657 ml=0.657 L

`\tt M=(0.504)/(0.657)\\\\M=0.767`