Question

EUWE How many grams of PbBr2 will precipitate when excess NaBr solution is added to 72.0 mL of 0.772 M Pb(NO3), solution? Pb(
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Answer #1

Answer

20.40g

Explanation

Pb(NO3)2(aq) + 2NaBr(aq) ------> PbBr2(s) + 2NaNO3(aq)

stoichiometrically, 1mole of Pb(NO3)2 gives 1mole of PbBr2

molarity = number of moles of solute per liter of solution

moles of Pb(NO3)2 = (0.772mol/1000ml)× 72ml = 0.055584mol

moles of PbBr2 precipitated = 0.055584mol

mass of PbBr2 precipitated = 0.055584mol × 367g/mol = 20.40g

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