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son4ous [18]
4 years ago
14

Iodine is prepared both in the laboratory and commercially by adding Cl2(g) to an aqueous solution containing sodium iodide: 2Na

I(aq) +Cl2(g) → I2(s) +2NaCl(aq) How many grams of iodide, NaI, must be used to produce 55.6 g of iodine, I2?
Chemistry
1 answer:
Oxana [17]4 years ago
8 0

Answer: 65.7 gram

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} I_2=\frac{55.6g}{254g/mol}=0.219moles

2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)

According to stoichiometry :

1 mole of I_2 are produced by = 2 moles of NaI

Thus 0.219 moles of I_2 will be produced by =\frac{2}{1}\times 0.219=0.438moles of NaI

Mass of NaI=moles\times {\text {Molar mass}}=0.438moles\times 150g/mol=65.7g

Thus 65.7 g of NaI, must be used to produce 55.6 g of iodine

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