According to the following reaction, how many grams of sodium chloride will be formed upon the complete reaction of 25.6 grams of sodium iodide with excess chlorine gas? chlorine (g) + sodium iodide (s) > sodium chloride (s) + iodine (s)

9.99 grams of NaCl

Explanation:

First, it has to be considered that the elemental form of halogens (like chlorine and iodine) are diatomic molecules. So, we can re-write the reaction as follows:

Cl2(g) + NaI(s) --> NaCl(s) + I2(s)

And then, if we balance the equation we will get the following:

Cl2(g) + 2NaI(s) --> 2NaCl(s) + I2(s)

So if we have enough Cl2(g) (as stated by the question in the part: "with excess chlorine gas"), 2 moles of NaI will produce 2 moles of NaCl. Considering the molar mass of NaI: 149.9 g/mol (Na=23, I=126.9) and applying a rule of three we can find out how many moles we have of NaI in 25.6 grams:

149.9 g --> 1 mol

25.6 g --> ? mol

? = 1.708x10^{-1} mol

So, if 2 moles of NaI produces 2 moles of NaCl, then 1.708x10^{-1} mol of NaI will produce 1.708x10^{-1} mol of NaCl

Now, we need to convert the moles of NaCl to grams using its molar mass: 58.5 g/mol (Na=23, Cl=35.5) and, as we did with NaI, we can apply a rule of three:

58.5 g --> 1 mol

? g --> 1.708x10^{-1} mol

? = 9.99 grams of NaCl

So, with the reaction of 25.6 grams of NaI (sodium iodide) with enough Cl2(g) (chlorine gas) we get 9.99 grams of NaCl (sodium chloride).

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