Qualitative analysis

Question

Calculate the mass of pure NaCl that will yield enough hydrogen chloride gas to neutralise 25 cm3 of 0.5M Potassium trioxocarbonate (iv) solution 

Answer 1

So like in the other question you posted, trioxocarbonate is simply another name for the Carbonate anion.  Cubic CM is equal to one mL.

 

A very general formula for this is the following:

 

NaCl(s) + (source of Hydrogen Ions probably NaHSO4(aq))  >>>> HCL (g) + K2CO3 (aq)

 

We just have to work backwards assuming 100% yield of products from reactants.

 

There is 0.5Mol/1Liter K2CO3 by definition of Molarity.

 

In only .025L of solution, that is .5/4= .125 moles of Carbonate neutralized.  With a 1:1 molar ratio of products, that means we have .125 moles of HCL that has reacted.

 

Assuming 100% yield, we can compare products to the limiting reagent (NaCl).  One again we have a 1:1 ratio.

 

So we have .125 moles of NaCl reacted.

 

The molar mass of NaCl is 58.439g/mol.

 

.125mol (58.439g/mol) = 7.3 grams of pure NaCl needed