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9.42 g of lead (ii) nitrate was reacted with 0.655 mol/L potassium iodide to produce 2.64 g of precipitate. What volume of potassium iodide was used in this reaction?

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Oseni

The volume of potassium iodide used will be 0.087 L or 87 mL

Stoichiometric calculation

The equation of the reaction goes thus: [tex]Pb(NO_3)_2(aq) + 2KI --- > PbI_2(s) + 2KNO_3(aq)[/tex]

The mole ratio of the 2 reactants is 1:2.

Mole of 9.42 g [tex]Pb(NO_3)_2[/tex] = 9.42/331.2 = 0.02844 mol

Equivalent mole of KI = 0.02844 x 2 = 0.05688 mol

Volume of 0.05688 mol, 0.655 mol/L KI = 0.05688/0.655 = 0.087 L or 87 mL

More on stoichiometric problems can be found here: https://brainly.com/question/15047541

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