Question

The mass of sodium chloride formed, when 5.3gm of sodium carbonate is dissolved in 250 ml of 0.5 molar HCL solution will be:-

Ans given:-5.85gm

Answer 1

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

From balanced equation, we have 2 mole HCl reacts with 1 mole Na2CO3 to give 2 mole NaCl.

Now, from given data,
Given mass of Na2CO3 = 5.3 grams
Molar mass of Na2CO3 = 106g
Hence no of moles = given mass / molar mass = 5.3/106= 0.05 moles of Na2CO3.

Now, for HCL,
molarity =M=0.5M.
By definition, molarity is no of moles in 1L solution.
So, HCl has 0.5 moles on 1L.
But we have only 250ml = 0.25 L.
Hence no of moles in 250 ml = 0.5 * 0.25 = 0.125 moles

So, we have 0.125 mole HCl and 0.05 mole Na2CO3.

From balanced equation above,
2 mole HCl needs 1 mole Na2CO3.
hence , 0.125 mole HCl needs 0.125/2 = 0.0625 moles Na2CO3. But we only have 0.05 moles Na2CO3.

So, Na2CO3 is limiting reagent , meaning, HCl doesn't completely react, but Na2CO3 completely reacts.

So , from balanced reaction, 1 mole Na2CO3 gives 2 mole NaCl.
So, 0.05 mole Na2CO3 gives 0.05*2 = 0.1 mole NaCl.

so, NaCl produced = 0.1 moles.
Mass of 1 mole NaCl = 58.5 grams.
So mass of produced 0.1 moles = 0.1 * 58.5 = 5.85 grams.