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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?

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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. :)

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