It takes 12.45 milliliters of 0.50 M NAOH to back titrate a mixture containing 1.25 grams of baking soda, NaHCO3, and 10.00 milliliters of 1.2 M HCL solution . Calculate the moles of antacid activity per gram of baking soda.

Explanation:

First, we will calculate the number of moles of both HCl and NaOH as follows.

     No. of moles of HCl = Molarity × Volume

                                      =        (as 1 L = 1000 ml)

                                      = 0.012 mol

   No. of moles of NaOH =  Molarity × Volume

                                    =      (12.45 ml = 0.1245 L)

                                   = 0.006 mol

Therefore, the amount of HCl reacts with will be as follows.

                   (0.012 - 0.06) mol

                  = 0.006 mol

Reaction equation for this reaction is as follows.

Here, one mole of HCl reacts with one mole of and it forms one mole of . Hence, moles of is 0.006 mol.

So,      

Here also, i mole of gives one mole of  .

So, = 0.006 moles

Now, moles of present per gram will be calculated as follows.

      Moles of per gram =

                              = 0.0048 g/mol

Thus, we can conclude that the moles of antacid activity per gram of baking soda is 0.0048 g/mol.

the answer is c i believe

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).