Question

10.6 g of pure sodium carbonate if treated with 7.9 g of HCl to produce NaCl, water and carbon dioxide. The mass of NaCl formed is

Answer 1

Answer:

Explanation: Na2CO3 + 2HCl -------> 2NaCl + H20

Here hcl is limiting reagent.

By mass mass relation,  73g HCl gives 117g  NaCl

                                    so 1g--------------->117/73g

                                    so 7.9g-------------> 117/73 *7.9 =12.66 gNaCl

So, ans is 12.66 g NaCl is produced.

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