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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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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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

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.