Intriguing456
contestada

You have 1.249 g of a mixture of NaHCO3 and Na2CO3. You find that 12.0 mL of 1.50 M HCl is required to convert the sample completely to NaCl, H2O, and CO2. How many moles of each compound in the mixture are there?

reaction equations:
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

Respuesta :

sqdancefan

Answer:

  • 0.009516 moles NaHCO₃
  • 0.004241 moles of Na₂CO₃

Explanation:

Given 1.249 g of a mixture of sodium bicarbonate and sodium carbonate reacts with 12.0 mL of 1.50 M HCl to convert the sample completely to NaCl, H₂O, and CO₂, you want to know the number of moles of each in the mix.

Molar masses

A web search finds the molar masses to be ...

  • NaHCO₃ — 84.007 g/mol
  • Na₂CO₃ — 105.9888 g/mol

Quantities

The number of moles of Cl involved in the reactions is ...

  0.012 L × 1.50 mol/L = 0.018 mol

Letting x and y represent the numbers of moles of NaHCO₃ and Na₂CO₃, respectively, we can write the equations ...

  84.007x +105.9888y = 1.249 . . . . . . . grams of reactants

  x + 2y = 0.018 . . . . . . . . . . . . . . . . . moles of Cl involved

The attachment shows the solution to these equations to be ...

  • 0.009516 moles NaHCO₃
  • 0.004241 moles of Na₂CO₃

__

Additional comment

There will be x + y ≈ 0.013758 moles each of water and CO₂ produced.

Ver imagen sqdancefan

Otras preguntas