Question

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 87.9 g of iodine, I2?

Answer 1

`m_(NaI)=104gNaI`

Step-by-step explanation:

Hello there!

In this case, given the balanced chemical reaction, it is possible to evidence that the mole ratio of sodium iodide to iodine is 2:1 and the molar masses are 149.89 and 253.81 g/mol respectively; in such a way, we write the following stoichiometric setup in order to obtain the required grams of sodium iodide:

`m_(NaI)=87.9gI_2*(1molI_2)/(253.81gI_2)*(2molNaI)/(1molI_2)*(149.89gNaI)/(1molNaI) \\\\m_(NaI)=104gNaI`

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