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27.80 grams of sodium iodide react with 5.20 grams of chlorine. When this experiment was carried out in the lab, the actual yield of sodium chloride was 6.55 grams. What is the remaining amount of reactants?

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When sodium iodide (NaI) reacts with chlorine gas, the following reaction takes place:

`2NaI (aq) + Cl_2 (g) -> 2NaCl (aq) + I_2 (aq)`

The products of this reaction are sodium chloride and iodine. 

Using stoichiometry, 2 moles of sodium iodide react with 1 mole of chlorine gas to produce 2 moles of sodium chloride and 1 mole of iodine.

The molar masses of all the species are:

sodium iodide = 150 g

chlorine gas = 71 g

sodium chloride = 58.5 g

iodine = 254 g

In this question, the masses of reactants are 27.8 g sodium iodide and 5.2 g chlorine gas and the mass of sodium chloride is given as 6.55 g. Let us convert all of them to number of moles of each species.

Sodium iodide = mass/molar mass = 27.8 g / 150 g/mole = 0.1853 moles

Chlorine gas = 5.2 g / 71 g/mole = 0.0732 moles

sodium chloride = 6.55 g / 58.5 g/mole = 0.112 moles.

Using stoichiometry, 2 moles of sodium iodide produce 2 moles of sodium chloride; however, we see that sodium iodide did not produce 0.1853 moles of sodium chloride. Similarly, chlorine gas also did not produce the requisite moles of sodium chloride and hence both sodium iodide and chlorine are left over.

To produce 0.112 moles sodium chloride, we need 0.112 moles of sodium iodide or 16.8 g (= 0.112 moles x 150 g/mole) of sodium iodide. Hence, 11 g sodium iodide would be left over. 

Similarly, 0.058 moles of chlorine would produce 0.112 moles of sodium iodide; thus, 0.0152 moles (= 0.0732 - 0.058) of chlorine is left over. We can also say that, 1.079 g (= 0.152 moles x 71 g/mole) of chlorine is left over.

So, to produce the given amount of sodium chloride, some portion, both of sodium iodide and chlorine, is left over. 

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