Question

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)

Part A

What volume of 1.00 M HCl in liters is needed to react completely (with nothing left over) with 0.500 L of 0.500 M Na2CO3?

Express your answer numerically in liters.

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nothing

L
L


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Part B

A 413-mL sample of unknown HCl solution reacts completely with Na2CO3 to form 11.1 g CO2. What was the concentration of the HCl solution?

Express the molar concentration numerically.

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nothing

M
M

Answer 1

Answer:

Part A: 0.500 L

Part B: 1.220 M

Explanation:

Part A

The balanced reaction is:

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

So, the stoichiometry is 2 moles of HCl for 1 mol of Na₂CO₃. The number of moles (n) is the concentration multiplied by the volume, so:

n = 0.500x0.500 = 0.250 mol of Na₂CO₃.

So:

2 moles of HCl ----------------- 1 mol of Na₂CO₃

      x                 ----------------- 0.250 mol of Na₂CO₃

By a simple direct three rule:

x = 0.500 mol

The volume will be the number of moles divided by the molar concentration:

V = 0.500/1.00 = 0.500 L

Part B

Now, the stoichiometry is 2 moles of HCl to form 1 mol of CO₂. We need to know how much moles of CO₂ are formed, knowing the molar masses:

MC = 12 g/mol

MO = 16 g/mol

MCO₂ = 12 +2x16 = 44 g/mol

The number of moles is the mass divided by the molar mass, so:

n = 11.1/44 = 0.252 mol

Then:

2 moles of HCl -------------------- 1 mol of CO₂

         x              -------------------- 0.252 mol of CO₂

By a simple direct three rule:

x = 0.504 mol of HCl

So, the concentration of HCl is the number of moles divided by the volume in liters ( 413 mL = 0.413 L):

0.504/0.413 = 1.220 M