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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reaction yield
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