Question

Solution Stoichiometry

Hydrochloric acid (HCl) reacts with sodium carbonate (Na2CO3), forming sodium chloride (NaCl), water (H2O), and carbon dioxide (CO2). This equation is balanced as written:

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

a)

What volume of 2.75 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.300 M Na2CO3?

b)

A 565-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 10.1 g CO2. What was the concentration of the HCl solution?

How do I solve these? Thanks

Answer 1

Answer-1

2HCl_(aq) + Na₂CO_3(aq) → 2NaCl_(aq) + H₂O_(l) + CO2_(g)

From above equation, we can say that two moles of HCl are required to neutralize Na2CO3.

We already know that, Na2CO3 is dissociated in water completely as Na+ and CO3-2 ions.

We also know that HCl is monobasic strong acid and ionizes in aqueous media in 1:1 ratio as below:

HCl_(aq) + H₂O_(l) → H3O+_(aq) + Cl-_(l)

So thecomplete ionic equation is 2 H+ (aq) + 2 Cl-_(l) + 2 Na+ (aq) + CO32- (aq) → H2CO3 + 2 Cl-_(l) + 2 Na+ (aq)

So, here Na= and Cl- ions are spectator ions. So Net Ionic reaction is

2 H+ (aq) + CO32- (aq) → H2CO3

c = n_solute / V_solution ⇒ n_solute = c⋅V_solution

Here,  

nCO₃^-2 = 0.300 mol L⁻¹ X 0.750L= 0.225 moles CO₃^-2

According to the above mentioned 1:2 mole ratio, a complete neutralization would require

0.225 moles CO₃^-2  X2 moles H₃O⁺ / 1 mole CO₃^-2 = 0.450 moles H₃O⁺

V_solution = n_solute / c

Therefore, V_H3O⁺= 0.450 moles / 2.75 mol L⁻¹ = 0.1636 L of 2.75 M HCl is required to to react completely (with nothing left over) with 0.750 L of 0.300 M Na2CO3.

Answer-2

Now, as discussed above we know that balance reaction is 2 HCl + Na2CO3 -----> CO2 + H2O + NaCl

Therefore,

moles of CO2  = weight / molecular weight

= 10.1 gm / 44 gm/mole = 0.22954

From balanced reaction we can say that 2 moles HCl react to form 1 mole of CO2 , therefore 0.22954 * 2 = 0.4591 moles HCl were reacted.

Now, moles= L x M

Therefore,

0.4591 moles HCl = 0.565 x M

M HCl = 00.4591 / 0.565

= 0.8126 M

So, the concentration of the HCl solution is 0.8126 M.