Question

1. Balance the reaction below and determine how many grams of nickel(II) phosphate will be formed when sodium phosphate (12.05g) is mixed with nickel(II) nitrate (175 ml, 0.500 M). After the reaction is complete, what is the concentration of the unprecipitated ion (either PO or Ni²+). Na,PO + Ni(NO3)2 - Ni,(PO),+ + NANO,

Answer 1

The complete solution is given in the following image:

164 2 Naz POq + 3 Ni (NO3)2 → Niz (Poa) + 6 NaNO3 from the equation we can see that 2 moles of Naz PO, leads to 1 mole of Niz (PO4)2 molar mass of Nazpoq = 169 g/mol molar mass of Niz (PO4)2 = 366 g/mol 12.85g of Nazpoq = 12:05 mol = 0.073 mol 175 mL of 0.500M Ni (103) = 173x10 % x 0.500 mol/l - 0.0865 mo/ So from 3 mol Ni(NO3), we get 1 mol Niz (809) hence from 0.0865 mol Ni(NO3), we get 1/2x0.08,65 mol = 0.029 mol Niz (P09), = 0.029 × 366= 10.6199 of Nizpod will be Ni (NOB) is the limiting reagent here. w Ni(103), we get į *°** Nig(P09). .. 0.029 mol = 0.029 x 3663) formed. (Answer)

1 mol of Nazpog contains 1 mol PO3- hence initial amount of pop? = Initial amount of Naz Pog = 0.073 mol I used Pog² = amount of Poq3 in product now I mol Niz(100% contains 2 mol Poq 3- used Pogle = 2x 0.029 mol = 0.058 mo) : amount of Poq3 in unprecipitated solution T S 10.073 – 0.058) = 0.015 mol and the volume of the solution = 175mL & Concentration of pog3- = 0.015mo/ of poq3 = 176x1032 = 0.085 mol/L