Iodine is prepared both in the laboratory and commercially by adding Cl2(g)to an aqueous solution containing sodium iodide.2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl( aq)How many grams of sodium iodide, NaI, must be used to produce 48.8 g of iodine, I2?

Hello there!

As you know, the products and reactants in a chemical reaction must be equal. Let's review the mole ratio for sodium iodide and iodine:

First of all, what even is a mole ratio? A mole ratio is the proportion between two substances. It is expressed using a colon in most instances, but can be expressed in fraction form.In this reaction, there are two moles of NaI. There is only one mole of iodine. Without any masses, this would look like the following:

                                       x      =     y

                                             

                                       2              1

Let's add the given mass for iodine in now:

                                       x      =     48.8g

                                             

                                       2              1

Now, using what we learned in math, let's solve this proportion:

                    1x = (48.8g *2)

                    1x = 97.6g

Seeing that any number divided by 1 is equal to itself, we can reasonably infer that 97.6 grams of NaI must be used to produce 48.7 of iodine.

I hope I helped. :)

A.(nh4)(2co3)is the correct answer