Question

Sodium carbonate reacts with hydrochloric acid to form sodium chloride water and carbon dioxide . Calculate mass of na2co3 required to completely react with 50ml 0.5m hcl

Answer 1

Let's first see the balanced reaction;

From this we can know that,

1 mole of Na2CO3 requires/reacts withs 2 moles of HCl to produce 2 moles of NaCl

We have, 5.3g of Na2CO3. Let's find the number of moles of Na2CO3 we have.

We also have 250ml of half molar HCl (Molarity = 0.5M). Let's find the number of moles of HCl.

So, finally, we have 0.05 moles of Na2CO3 and 0.25 moles of HCl.

If 1 mole of Na2CO3 reacts with 2 moles of HCl to give products,

0.05 miles of Na2CO3 will react with (2 × 0.05 =) 0.1 moles of HCl.

But we have more than the required amount of HCl, which means some amount of HCl is remain even after the reaction terminates. Whereas, Na2CO3 will be completely consumed during the reactions. So that is the Limiting Reagent.

Since, Na2CO3 is the Limiting Reagent, the amount of product produced will depend on its concentration.

If 1 mole of Na2CO3 reacts with HCl to give 2 moles of NaCl,

0.05 mole of Na2CO3 reacts with HCl to give (2 × 0.05 =) 0.10 moles of NaCl.

Therefore, 5.85g of NaCl will be produced when 5.3g of Na2CO3 reacts with 250ml of 0.5M HCl solution.

Thanks.