Question

Solution Stoichiometry Hydrochloric acid (HCl) reacts with sodium carbonate (Na2CO3), forming sodium chloride (NaCl), water (H2O),...

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

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Answer #1

Answer-1

2HCl(aq) + Na2CO3(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 = nsolute / Vsolution ⇒ nsolute = c⋅Vsolution

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+

Vsolution = nsolute / c

Therefore, VH3O+= 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.

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