Question

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. (a) Write a balanced equation for each step. (b) How many grams of iodine are needed to prepare 2.45 kg of final product, assuming 100% yield for each step? (c) How many grams of iodine are needed to prepare 2.45 kg of final product, assuming 80% yield for each step?

Answer 1

step 1)    I₂ + Cl₂ → 2 ICl

step 2)   ICl + Cl₂ → ICl₃

mass of I₂ = 1333.5 g (100% yield)

practical mass of I₂ = 853.44 g (64% yield)

Explanation:

(a) Write a balanced equation for each step.

step 1)    I₂ + Cl₂ → 2 ICl

step 2)   ICl + Cl₂ → ICl₃

global reaction     I₂ + 3 Cl₂ → 2 ICl₃

(b) How many grams of iodine are needed to prepare 2.45 kg of final product, assuming 100% yield for each step?

molecular mass of ICl₃ = 233.5 g/mol

number of moles = mass / molecular mass

number of moles of ICl₃ = 2450 / 233.5

number of moles of ICl₃ = 10.49 moles

From the global reaction we see that 1 mole of iodine I₂  produces 2 moles of 2 ICl₃.

if        1 mole of I₂  produces 2 moles of 2 ICl₃

then  X moles of I₂  produces 10.49 moles of 2 ICl₃

X = (1 × 10.49) / 2 = 5.25 moles of I₂ (100% yield)

mass = number of moles × molecular mass

mass of I₂ = 5.25 × 254

mass of I₂ = 1333.5 g

(c) How many grams of iodine are needed to prepare 2.45 kg of final product, assuming 80% yield for each step?

yield of the global reaction = yield of step (1) × yield of step (1)

yield of the global reaction = (80/100) × (80/100) = 64/100 = 64 %

yield = (practical mass / theoretical mass) × 100

practical mass = (yield × theoretical mass) / 100

practical mass of I₂ = (64 × 1333.5) / 100

practical mass of I₂ = 853.44 g

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