Question

How many mL of a 9.4% by mass ammonia solution (D=1.32 gxmL^-1) are required to react completely with 105g of lead (II) cyanate solid?

Pb(NCO)2 (s)+ 2NH3 (aq) + H2O (l) ------> 2NH4(NCO) (aq) + Pb(OH)2 (aq) (s)

*Please explain in detail*

Answer 1

molar mass of Pb(CNO)2 = 1* atomic weight of lead +2*( atomic weight of C+ atomic weigh ot N+ atomic weight of O)

=207.2+2*(12+14+16) =291.2

moles of lead cynate= mass/molar mass= 105/291.2= 0.36

as per the reaction,

Pb(NCO)2 (s)+ 2NH3 (aq) + H2O (l) ------> 2NH4(NCO) (aq) + Pb(OH)2 (aq) (s)

1 mole of Pb(NCO)2 requires 2moles of NH3

hence 0.36 moles of Pb(NCO)2 requires 2*0.36=0.72 moles of NH3

let xml of solution is used. mass of solution = volume in ml* density= x*1.32= 1.32x

the solution contains 9.4% of NH3, mass of NH3 in the solution = mass of solution*9.4/100= 1.32*9.4x/100 = 0.12x

moles of NH3 = mass of NH3/ molar mass= 0.12x/(14+3*1)= 0.12x/17=0.0071x

but 0.0071x=0.72

x= 101.4 ml