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

yes

explanation:

search chemistry final b exam in google and a site called quizlet will pop up

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.