Calcullate the number of moles of chlorine that are needed to make 6.00 g of lead (IV)
chloride. [0.0344 mol Cl,]
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Moles of Cl₂ = 0.0344
Given
6.00 g of Lead (IV) chloride
Required
Moles of Chlorine
Solution
Reaction
Balanced equation :
Pb + 2Cl₂⇒ PbCl₄
mol of PbCl₄(MW= 349.012 g/mol[) :
mol = mass : MW
mol = 6 g : 349.012 g/mol
mol = 0.0172
From the equation, mol ratio of Cl₂ : PbCl₄ = 2 : 1, so mol Cl₂ :
= 2 x 0.0172
= 0.0344