Question

For the reaction 2 KI + Pb(NO), — Pol, + 2 KNO, how many grams of lead(II) nitrate, Pb(NO3)2, are needed to react completely with 39.3 g of potassium iodide, KI? mass:

Answer 1

1)

Molar mass of KI,

MM = 1*MM(K) + 1*MM(I)

= 1*39.1 + 1*126.9

= 166 g/mol

mass of KI = 39.3 g

mol of KI = (mass)/(molar mass)

= 39.3/1.66*10^2

= 0.2367 mol

According to balanced equation

mol of Pb(NO3)2 formed = (1/2)* moles of KI

= (1/2)*0.2367

= 0.1184 mol

Molar mass of Pb(NO3)2,

MM = 1*MM(Pb) + 2*MM(N) + 6*MM(O)

= 1*207.2 + 2*14.01 + 6*16.0

= 331.22 g/mol

mass of Pb(NO3)2 = number of mol * molar mass

= 0.1184*3.312*10^2

= 39.21 g

Answer: 39.2 g

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