Question

Aqueous silver nitrate reacts with aqueous lead (IV) acetate. Calculate the grams of each product formed, starting witb 2.001g of silver nitrate.

Answer 1

Given the mass of AgNO3 taken = 2.001 g

Molecular mass of AgNO3 = 169.87 gmol^-1

Hence moles of AgNO3 taken = mass / molecular mass = 2.001g / 169.87 gmol^-1 = 0.01178 mol

The balanced chemical equation for the reaction of silver nitrate(AgNO3) with lead(IV) acetate is

4AgNO3(aq) + Pb(CH3COO)4(aq) -- > 4 CH3COOAg(s) + Pb(NO3)4 (aq)

4 mol 1 mol 4 mol 1 mol

For the above balanced chemical reaction,

4 mol of AgNO3 forms 4 mol of CH3COOAg.

Hence 0.01178 mol of AgNO3 that will form the moles of CH3COOAg

= (0.01178 mol AgNO3) x (4 mol CH3COOAg / 4 mol AgNO3) = 0.01178 mol CH3COOAg(s)

Molecular mass of CH3COOAg(s) = 166.91 gmol^-1

Hence mass of CH3COOAg(s) = 0.01178 mol x166.91 gmol^-1 = 1.966 g CH3COOAg(s) (answer)

4 mol of AgNO3 forms 1 mol of Pb(NO3)4 (aq)

Hence 0.01178 mol of AgNO3 that will form the moles of Pb(NO3)4 (aq)

= (0.01178 mol AgNO3) x (1 mol Pb(NO3)4 (aq) / 4 mol AgNO3) = 0.002945 mol Pb(NO3)4 (aq)

Molecular mass of Pb(NO3)4 (aq) = 455.22 gmol^-1

Hence mass of Pb(NO3)4 (aq) formed = 0.002945 mol x455.22 gmol^-1 = 1.341 g Pb(NO3)4 (aq) (answer)