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).

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.