Answer to Question #209725 in Chemistry for angie

Question

225 mL of a 1.45 mol/L solution of sodium iodide is added to 175 mL of a 0.85 mol/L solution of lead(II) nitrate

a) write the balanced chemical equation for this reaction

b) how many grams of solid precipitate will form in the process

c) what is the percent yield if 57.8g of precipitate was collected?

Answer 1

a) 2NaI + Pb(NO₃)₂ = PbI₂ + 2NaNO₃

b) CM = n / V

n = C_M x V

n (NaI) = 1.45 x 0.225 = 0.33 mol

n (Pb(NO₃)₂) = 0.85 x 0.175 = 0.15 mol

From the equation, n (NaI) = 2 x n (Pb(NO₃)₂).

Therefore, in this reaction Pb(NO₃)₂ is the limiting reactant.

n (PbI₂) = n (Pb(NO₃)₂) = 0.15 mol

M (PbI₂) = 461.01 g/mol

n = m / M

m (PbI₂) = n x M = 0.15 x 461.01 = 69.15 g

c) w (PbI₂) = 57.8 / 69.15 x 100 = 83.59%