Calcullate the number of moles of chlorine that are needed to make 6.00 g of lead (IV) chloride. [0.0344 mol Cl,]

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

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