Problem: Solid iodine trichloride is prepared in two steps: first, a reaction between solid iodine and gaseous chlorine to form solid iodine monochloride; second, treatment of the solid with more chlorine gas. (c) How many grams of iodine are needed to prepare 2.45 kg of final product?

🤓 Based on our data, we think this question is relevant for Professor Veige's class at UF.

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1st reaction:                         solid iodine + gaseous chlorine → solid iodine monochloride

Balanced reaction:               I2(s) + Cl2(g) → 2 ICl(s)

2nd reaction:                         treatment of the solid with more chlorine gas

Balanced reaction:               2 ICl(s) + 2 Cl2(g) → 2 ICl3(s)


1 mol of iodine produces 2 moles of ICl and 2 moles of ICl produces to moles of ICl3

molar mass ICl3 = 233.25 g/mol

molar mass I2 = 253.8 g/mol

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Problem Details

Solid iodine trichloride is prepared in two steps: first, a reaction between solid iodine and gaseous chlorine to form solid iodine monochloride; second, treatment of the solid with more chlorine gas. 

(c) How many grams of iodine are needed to prepare 2.45 kg of final product?

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Our tutors have indicated that to solve this problem you will need to apply the Stoichiometry concept. You can view video lessons to learn Stoichiometry. Or if you need more Stoichiometry practice, you can also practice Stoichiometry practice problems.

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Based on our data, we think this problem is relevant for Professor Veige's class at UF.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.